---
title: "The Mole and Molar Mass"
book: "School Chemistry — Grades 1 to 12"
subject: chemistry
language: en
chapter: 25
exercises: 15
source: https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 25 — The Mole and Molar Mass

At the end of the day, a bank does not count its coins one by one: it weighs them. If one coin has a mass of $7.5\,\mathrm{g}$, a bag of coins with a mass of $7.5\,\mathrm{kg}$ holds a thousand coins, and the scale has done the counting. Chemists face the same problem with [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), only far worse: a spoonful of water holds more [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) than there are grains of sand on all the beaches of a coast. They solve it the same way, by weighing, and the bag they use holds $602\,214\,076\,000\,000\,000\,000\,000$ entities. This chapter defines that bag, the [mole](#def-g10-the-mole-amount), and shows how to go from a mass on a balance to a number of [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) or [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule).

**You already know.**

The mass of an [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) is concentrated in its [nucleus](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus); a [proton](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-proton) and a [neutron](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-proton) each have a mass of about $1.67 \times 10^{-27}\,\mathrm{kg}$, so an [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of [mass number](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-atomic-number) $A$ has a mass of about $A \times 1.67 \times 10^{-27}\,\mathrm{kg}$. Large and small numbers are written in scientific notation ([Chapter 16](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#ch-g9-inside-the-atom)). A [chemical formula](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-formula) gives the number of [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of each element in a [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) ([Chapter 11](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#ch-g7-atoms-and-molecules)).

![A bank counts coins by weighing them. Chemists count atoms the same way.](https://one-course.com/images/onecourse/chapters/chemistry-1/g10-the-mole/img-d368192ff5cf.jpg)

*A bank counts coins by weighing them. Chemists count [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) the same way.*

## 25.1 Counting by weighing

**Example 25.1 (A bag of coins).**

A coin has a mass of $7.5\,\mathrm{g}$. A bag of these coins has a mass of $1.80\,\mathrm{kg}$, that is $1800\,\mathrm{g}$, the mass of the bag itself left aside. It holds

$$
\frac{1800\,\mathrm{g}}{7.5\,\mathrm{g}} = 240 \text{ coins.}
$$

To count, the bank divides the total mass by the mass of one coin.

**Example 25.2 (A spoonful of water).**

A water [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule), $\ce{H2O}$, has $2 + 16 = 18$ [nucleons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-proton) (the oxygen-16 and hydrogen-1 [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) make up nearly all natural water), so its mass is about $18 \times 1.67 \times 10^{-27}\,\mathrm{kg} = 3.0 \times 10^{-26}\,\mathrm{kg}$, or $3.0 \times 10^{-23}\,\mathrm{g}$. A spoonful of $5.0\,\mathrm{g}$ of water therefore holds

$$
\frac{5.0\,\mathrm{g}}{3.0 \times 10^{-23}\,\mathrm{g}} \approx 1.7 \times 10^{23}
  \text{ molecules.}
$$

The division works as for the coins, but the answer is a number no one can picture. Chemists need a package of [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) adapted to the samples of the laboratory.

**Definition 25.3 (Amount of substance and mole).**

The *amount of substance* $n$ of a sample counts the entities it contains ([atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) or [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), stated each time). Its unit is the *mole*, symbol $\mathrm{mol}$: one mole contains exactly $6.022\,140\,76 \times 10^{23}$ entities.

**Remark 25.4 (A dozen, a ream, a mole).**

The [mole](#def-g10-the-mole-amount) is a counting unit, like a dozen eggs (12) or a ream of paper (500 sheets), only very much larger. “Two [moles](#def-g10-the-mole-amount) of water [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule)” means $2 \times 6.022\,140\,76 \times 10^{23} = 1.20 \times 10^{24}$ [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule), just as “two dozen eggs” means 24 eggs. Always say which entities are counted: one [mole](#def-g10-the-mole-amount) of oxygen *[molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule)* $\ce{O2}$ holds two [moles](#def-g10-the-mole-amount) of oxygen *[atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom)*.

**History — Avogadro’s hypothesis, 1811.**

In 1811 the physicist Amedeo Avogadro proposed that equal volumes of different gases, at the same temperature and pressure, contain the same number of particles. He also understood that a gas such as oxygen is made of [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of two [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom). His idea was neglected for nearly fifty years, until Stanislao Cannizzaro used it in 1860 to obtain a coherent set of atomic weights. The constant that counts the entities of a [mole](#def-g10-the-mole-amount) was later named after Avogadro; he never knew its value.

![Amedeo Avogadro (1776–1856).](https://one-course.com/images/onecourse/chapters/chemistry-1/g10-the-mole/img-e2742f95a956.jpg)

*Amedeo Avogadro (1776–1856).*

## 25.2 The Avogadro constant

**Definition 25.5 (Avogadro constant).**

The *Avogadro constant* $N_A$ is the number of entities per [mole](#def-g10-the-mole-amount):

$$
N_A = 6.022\,140\,76 \times 10^{23}\,\mathrm{mol}^{-1} .
$$

Its value is exact, fixed by definition. In calculations it is rounded to $6.02 \times 10^{23}\,\mathrm{mol}^{-1}$.

**Proposition 25.6 (Number of entities and amount).**

A sample holding $N$ entities contains the amount

$$
n = \frac{N}{N_A}, \qquad\text{that is}\qquad N = n \times N_A .
$$

**Proof.** Each [mole](#def-g10-the-mole-amount) contains $N_A$ entities, so $n$ [moles](#def-g10-the-mole-amount) contain $n \times N_A$ entities; dividing by $N_A$ gives $n$. ∎

**Example 25.7 (Molecules and moles).**

A sample contains $1.5 \times 10^{24}$ [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of carbon dioxide. Its amount of carbon dioxide is $n = 1.5 \times 10^{24} / 6.02 \times 10^{23}\,\mathrm{mol}^{-1} = 2.5\,\mathrm{mol}$. Conversely, $0.20\,\mathrm{mol}$ of iron holds $N = 0.20\,\mathrm{mol} \times 6.02 \times 10^{23}\,\mathrm{mol}^{-1} = 1.2 \times 10^{23}$ [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of iron.

**Remark 25.8 (Where the number comes from).**

For more than fifty years the [mole](#def-g10-the-mole-amount) was defined as the number of [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) in exactly $12\,\mathrm{g}$ of carbon-12, and $N_A$ had to be measured. Since 2019 the number itself has been fixed, which makes the [mole](#def-g10-the-mole-amount) a pure count. The old and new definitions agree to better than one part in a hundred million, so twelve grams of carbon-12 still contain one [mole](#def-g10-the-mole-amount) of [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), as the next section checks.

## 25.3 Molar mass

**Definition 25.9 (Molar mass).**

The *molar mass* $M$ of a [chemical species](https://one-course.com/books/chemistry/1/en/chapter/8-pure-substances-and-mixtures#def-g6-pure-substances-and-mixtures-species) is the mass of one [mole](#def-g10-the-mole-amount) of its entities, in grams per [mole](#def-g10-the-mole-amount) ($\mathrm{g}/\mathrm{mol}$). The molar mass of an element, taken over its natural [mixture](https://one-course.com/books/chemistry/1/en/chapter/8-pure-substances-and-mixtures#def-g6-pure-substances-and-mixtures-mixture) of [isotopes](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-isotope), is its *atomic molar mass*; it is printed in the [periodic table](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-periodic-table).

**Example 25.10 (One mole of carbon atoms).**

An [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of carbon-12 has a mass of about $12 \times 1.67 \times 10^{-27}\,\mathrm{kg} = 2.00 \times 10^{-26}\,\mathrm{kg}$. One [mole](#def-g10-the-mole-amount) of them has a mass of

$$
2.00 \times 10^{-26}\,\mathrm{kg} \times 6.02 \times 10^{23} = 1.20 \times 10^{-2}\,\mathrm{kg}
  = 12.0\,\mathrm{g}.
$$

The [mole](#def-g10-the-mole-amount) has been chosen so that the [molar mass](#def-g10-the-mole-molar-mass) of an [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), in grams per [mole](#def-g10-the-mole-amount), is close to its [mass number](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-atomic-number): about $1\,\mathrm{g}/\mathrm{mol}$ for hydrogen, $12\,\mathrm{g}/\mathrm{mol}$ for carbon, $16\,\mathrm{g}/\mathrm{mol}$ for oxygen.

| element | symbol | $M$ ($\mathrm{g}/\mathrm{mol}$) | element | symbol | $M$ ($\mathrm{g}/\mathrm{mol}$) |
| --- | --- | --- | --- | --- | --- |
| hydrogen | $\ce{H}$ | 1.0 | silicon | $\ce{Si}$ | 28.1 |
| carbon | $\ce{C}$ | 12.0 | sulfur | $\ce{S}$ | 32.1 |
| nitrogen | $\ce{N}$ | 14.0 | chlorine | $\ce{Cl}$ | 35.5 |
| oxygen | $\ce{O}$ | 16.0 | potassium | $\ce{K}$ | 39.1 |
| sodium | $\ce{Na}$ | 23.0 | calcium | $\ce{Ca}$ | 40.1 |
| magnesium | $\ce{Mg}$ | 24.3 | iron | $\ce{Fe}$ | 55.8 |
| aluminium | $\ce{Al}$ | 27.0 | copper | $\ce{Cu}$ | 63.5 |
|  |  |  | gold | $\ce{Au}$ | 197.0 |

*Atomic molar masses of common elements, rounded to $0.1\,\mathrm{g}/\mathrm{mol}$. These are the values used in this book.*

**Remark 25.11 (Why chlorine is 35.5).**

Natural chlorine is a [mixture](https://one-course.com/books/chemistry/1/en/chapter/8-pure-substances-and-mixtures#def-g6-pure-substances-and-mixtures-mixture) of two [isotopes](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-isotope): about three quarters of its [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) are chlorine-35, one quarter chlorine-37 ([Chapter 16](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#ch-g9-inside-the-atom)). Its [atomic molar mass](#def-g10-the-mole-molar-mass) is the average, weighted by these shares, which lies between 35 and 37: $35.5\,\mathrm{g}/\mathrm{mol}$. That is why some atomic molar masses are far from whole numbers.

**Proposition 25.12 (Molar mass of a molecule).**

The [molar mass](#def-g10-the-mole-molar-mass) of a [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) is the sum of the atomic molar masses of all its [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), each counted as many times as it appears in the formula. The same rule gives the [molar mass](#def-g10-the-mole-molar-mass) of an ionic solid from its formula unit.

**Proof.** One [mole](#def-g10-the-mole-amount) of [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) $\ce{A_xB_y}$ contains $x$ [moles](#def-g10-the-mole-amount) of [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) $\ce{A}$ and $y$ [moles](#def-g10-the-mole-amount) of [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) $\ce{B}$, and nothing else; its mass is the sum of their masses, $x\,M(\ce{A}) + y\,M(\ce{B})$. ∎

**Example 25.13 (Three molar masses).**

$$
\begin{align*}
  M(\ce{H2O}) &= 2 \times 1.0 + 16.0 = 18.0\,\mathrm{g}/\mathrm{mol},\\
  M(\ce{NaCl}) &= 23.0 + 35.5 = 58.5\,\mathrm{g}/\mathrm{mol},\\
  M(\ce{C12H22O11}) &= 12 \times 12.0 + 22 \times 1.0 + 11 \times 16.0
     = 342.0\,\mathrm{g}/\mathrm{mol}.
\end{align*}
$$

The last one is sucrose, table sugar.

## 25.4 From mass to amount and back

**Proposition 25.14 (Mass and amount).**

A sample of mass $m$ of a species of [molar mass](#def-g10-the-mole-molar-mass) $M$ contains the amount

$$
n = \frac{m}{M}, \qquad\text{that is}\qquad m = n \times M .
$$

With $m$ in grams and $M$ in grams per [mole](#def-g10-the-mole-amount), $n$ is in [moles](#def-g10-the-mole-amount).

**Proof.** Each [mole](#def-g10-the-mole-amount) has a mass $M$, so $n$ [moles](#def-g10-the-mole-amount) have a mass $n \times M$; this is the coin count of the bank, with the [mole](#def-g10-the-mole-amount) as the coin. ∎

**Method 25.15 (From a mass to a number of entities).**

1. Write the formula of the species and compute its [molar mass](#def-g10-the-mole-molar-mass) $M$ from the table.
2. Convert the mass to grams if needed, then $n = m / M$ .
3. If the number of entities is wanted, $N = n \times N_A$ .
4. Check the order of magnitude: a few grams of a small [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) is of the order of a tenth of a [mole](#def-g10-the-mole-amount) , some $10^{22}$ to $10^{23}$ entities.

![The amount n sits in the middle: the balance gives m, the molar mass leads to n, and the Avogadro constant leads on to the number of entities.](https://one-course.com/images/onecourse/chapters/chemistry-1/g10-the-mole/fig-73f7143a0480.svg)

*The amount $n$ sits in the middle: the balance gives $m$, the [molar mass](#def-g10-the-mole-molar-mass) leads to $n$, and the [Avogadro constant](#def-g10-the-mole-avogadro) leads on to the number of entities.*

**Example 25.16 (Nine grams of water).**

How many [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) are there in $9.0\,\mathrm{g}$ of water? $M(\ce{H2O}) =
18.0\,\mathrm{g}/\mathrm{mol}$, so

$$
n = \frac{9.0\,\mathrm{g}}{18.0\,\mathrm{g}/\mathrm{mol}} = 0.50\,\mathrm{mol},
  \qquad
  N = 0.50\,\mathrm{mol} \times 6.02 \times 10^{23}\,\mathrm{mol}^{-1}
    = 3.0 \times 10^{23} \text{ molecules.}
$$

**Example 25.17 (A mole on the bench).**

One [mole](#def-g10-the-mole-amount) of water is $18.0\,\mathrm{g}$, about a tablespoon. One [mole](#def-g10-the-mole-amount) of salt, $58.5\,\mathrm{g}$, is a small heap; one [mole](#def-g10-the-mole-amount) of sugar, $342.0\,\mathrm{g}$, fills a mug; one [mole](#def-g10-the-mole-amount) of iron, $55.8\,\mathrm{g}$, is a cube a little under $2\,\mathrm{cm}$ across. All four hold the same number of entities: the [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of sugar are simply much heavier than those of water, and the [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of iron are packed much more tightly.

![One mole each of water, salt (sodium chloride), sugar (sucrose) and iron, drawn as cubes of their true volumes, all at the same scale. Each holds 6.02 × 1023 entities.](https://one-course.com/images/onecourse/chapters/chemistry-1/g10-the-mole/fig-bdbd5fadfef4.svg)

*One [mole](#def-g10-the-mole-amount) each of water, salt (sodium chloride), sugar (sucrose) and iron, drawn as cubes of their true volumes, all at the same scale. Each holds $6.02 \times 10^{23}$ entities.*

## 25.5 Gases: the molar volume

**Definition 25.18 (Molar volume).**

The *molar volume* $V_m$ of a gas is the volume occupied by one [mole](#def-g10-the-mole-amount) of that gas, at a given temperature and pressure. It is expressed in litres per [mole](#def-g10-the-mole-amount) ($\mathrm{L}/\mathrm{mol}$).

**Proposition 25.19 (All gases have the same molar volume).**

At a given temperature and pressure, one [mole](#def-g10-the-mole-amount) of any gas occupies very nearly the same volume, whatever the gas. At $20\,{}^{\circ}\mathrm{C}$ and normal atmospheric pressure,

$$
V_m = 24.1\,\mathrm{L}/\mathrm{mol};
$$

at $0\,{}^{\circ}\mathrm{C}$ and the same pressure, $V_m = 22.4\,\mathrm{L}/\mathrm{mol}$. The amount of a gas of volume $V$ is then $n = V / V_m$.

**Proof.** Admitted here: this is Avogadro’s hypothesis of 1811, now a law of physics, and the value of $V_m$ follows from the gas laws studied in the physics book. ∎

![Three balloons at 20\, C, each holding one mole of a different gas. The volumes are the same; the masses are not.](https://one-course.com/images/onecourse/chapters/chemistry-1/g10-the-mole/fig-fc6212e65490.svg)

*Three balloons at $20\,{}^{\circ}\mathrm{C}$, each holding one [mole](#def-g10-the-mole-amount) of a different gas. The volumes are the same; the masses are not.*

**Example 25.20 (A balloon of carbon dioxide).**

A balloon holds $2.4\,\mathrm{L}$ of carbon dioxide at $20\,{}^{\circ}\mathrm{C}$. Its amount is $n = 2.4\,\mathrm{L} / 24.1\,\mathrm{L}/\mathrm{mol} = 0.10\,\mathrm{mol}$, and its mass $m = 0.10\,\mathrm{mol} \times 44.0\,\mathrm{g}/\mathrm{mol} =
4.4\,\mathrm{g}$. The same balloon filled with hydrogen would hold the same $0.10\,\mathrm{mol}$, but only $0.20\,\mathrm{g}$ of gas.

**Remark 25.21 (Heavy and light gases).**

Since a litre of any gas holds the same amount, the mass of a litre of gas is proportional to its [molar mass](#def-g10-the-mole-molar-mass). [Air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air), a [mixture](https://one-course.com/books/chemistry/1/en/chapter/8-pure-substances-and-mixtures#def-g6-pure-substances-and-mixtures-mixture) of about four fifths nitrogen ($28.0\,\mathrm{g}/\mathrm{mol}$) and one fifth oxygen ($32.0\,\mathrm{g}/\mathrm{mol}$), behaves like a gas of [molar mass](#def-g10-the-mole-molar-mass) about $29\,\mathrm{g}/\mathrm{mol}$. A gas of larger [molar mass](#def-g10-the-mole-molar-mass), such as carbon dioxide ($44.0\,\mathrm{g}/\mathrm{mol}$), is denser than [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) and gathers at the bottom of a closed cellar; a gas of smaller [molar mass](#def-g10-the-mole-molar-mass), such as helium ($4.0\,\mathrm{g}/\mathrm{mol}$), rises.

## 25.6 Exercises

**Exercise 25.1 ★.**

Compute the molar masses of dioxygen $\ce{O2}$, dinitrogen $\ce{N2}$, methane $\ce{CH4}$, ammonia $\ce{NH3}$ and carbon dioxide $\ce{CO2}$.

**Solution of Exercise 25.1.**

$M(\ce{O2}) = 2 \times 16.0 = 32.0\,\mathrm{g}/\mathrm{mol}$; $M(\ce{N2}) = 2 \times 14.0 = 28.0\,\mathrm{g}/\mathrm{mol}$; $M(\ce{CH4}) = 12.0 + 4 \times 1.0 = 16.0\,\mathrm{g}/\mathrm{mol}$; $M(\ce{NH3}) = 14.0 + 3 \times 1.0 = 17.0\,\mathrm{g}/\mathrm{mol}$; $M(\ce{CO2}) = 12.0 + 2 \times 16.0 = 44.0\,\mathrm{g}/\mathrm{mol}$.

**Exercise 25.2 ★.**

What [amount of substance](#def-g10-the-mole-amount) is contained in $36.0\,\mathrm{g}$ of water? In $4.0\,\mathrm{g}$ of methane?

**Solution of Exercise 25.2.**

Water: $n = 36.0 / 18.0 = 2.00\,\mathrm{mol}$. Methane: $n = 4.0 / 16.0 = 0.25\,\mathrm{mol}$.

**Exercise 25.3 ★.**

What is the mass of $0.25\,\mathrm{mol}$ of sodium chloride $\ce{NaCl}$? Of $2.0\,\mathrm{mol}$ of carbon dioxide?

**Solution of Exercise 25.3.**

$m(\ce{NaCl}) = 0.25 \times 58.5 = 14.6\,\mathrm{g}$, about $15\,\mathrm{g}$; $m(\ce{CO2}) = 2.0 \times 44.0 = 88\,\mathrm{g}$.

**Exercise 25.4 ★.**

How many [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) are there in $0.50\,\mathrm{mol}$ of dioxygen? What amount of iron contains $3.0 \times 10^{24}$ [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of iron?

**Solution of Exercise 25.4.**

$N = 0.50 \times 6.02 \times 10^{23} = 3.0 \times 10^{23}$ [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of dioxygen. $n = 3.0 \times 10^{24} / 6.02 \times 10^{23}\,\mathrm{mol}^{-1} = 5.0\,\mathrm{mol}$ of iron.

**Exercise 25.5 ★.**

At $20\,{}^{\circ}\mathrm{C}$ and normal atmospheric pressure, what volume does $0.50\,\mathrm{mol}$ of dinitrogen occupy? What amount of dioxygen is there in $12\,\mathrm{L}$ of that gas?

**Solution of Exercise 25.5.**

$V = n \times V_m = 0.50 \times 24.1 = 12\,\mathrm{L}$. $n = V / V_m = 12 / 24.1 = 0.50\,\mathrm{mol}$.

**Exercise 25.6 ★★.**

A drop of water has a volume of $0.050\,\mathrm{mL}$, and $1\,\mathrm{mL}$ of water has a mass of $1.0\,\mathrm{g}$. How many water [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) does the drop contain?

**Solution of Exercise 25.6.**

The drop has a mass of $0.050\,\mathrm{g}$, so $n = 0.050 / 18.0 = 2.8 \times 10^{-3}\,\mathrm{mol}$ and $N = 2.8 \times 10^{-3} \times 6.02 \times 10^{23} = 1.7 \times 10^{21}$ [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule).

**Exercise 25.7 ★★.**

Which contains more [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), $10\,\mathrm{g}$ of aluminium or $10\,\mathrm{g}$ of iron? Explain without computing first, then check by computing.

**Solution of Exercise 25.7.**

An aluminium [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) ($27.0\,\mathrm{g}/\mathrm{mol}$) is about half as heavy as an iron [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) ($55.8\,\mathrm{g}/\mathrm{mol}$), so the same mass of aluminium holds about twice as many [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom). Check: aluminium $10/27.0 = 0.370\,\mathrm{mol}$, that is $2.23 \times 10^{23}$ [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom); iron $10/55.8 = 0.179\,\mathrm{mol}$, that is $1.08 \times 10^{23}$ [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom). Aluminium wins, by a factor of about 2.1.

**Exercise 25.8 ★★.**

At $20\,{}^{\circ}\mathrm{C}$, compute the mass of one litre of carbon dioxide and of one litre of dinitrogen. Why does carbon dioxide collect at the bottom of a closed cellar where grape juice is fermenting?

**Solution of Exercise 25.8.**

One litre is $1/24.1 = 0.0415\,\mathrm{mol}$ of gas. Carbon dioxide: $0.0415 \times 44.0 = 1.83\,\mathrm{g}$; dinitrogen: $0.0415 \times 28.0 = 1.16\,\mathrm{g}$. Carbon dioxide is about one and a half times denser than [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air), whose [molar mass](#def-g10-the-mole-molar-mass) is close to that of dinitrogen. The fermenting juice gives off carbon dioxide, which sinks and builds up near the floor of a closed cellar, where it can suffocate anyone who goes in.

**Exercise 25.9 ★★.**

Using the figure of the three boxes $m$, $n$, $N$, give the mass of $1.0 \times 10^{22}$ [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of glucose $\ce{C6H12O6}$. Write down the two arrows followed.

**Solution of Exercise 25.9.**

$M(\ce{C6H12O6}) = 6 \times 12.0 + 12 \times 1.0 + 6 \times 16.0 =
180.0\,\mathrm{g}/\mathrm{mol}$. Arrow $\div N_A$: $n = 1.0 \times 10^{22} / 6.02 \times 10^{23} =
1.66 \times 10^{-2}\,\mathrm{mol}$; arrow $\times M$: $m = 1.66 \times 10^{-2} \times 180.0 =
3.0\,\mathrm{g}$.

**Exercise 25.10 ★★.**

A lump of sugar (sucrose, $\ce{C12H22O11}$) has a mass of $5.0\,\mathrm{g}$. Compute the amount of sucrose it contains, then the number of sucrose [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule), then the number of carbon [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom).

**Solution of Exercise 25.10.**

$M = 342.0\,\mathrm{g}/\mathrm{mol}$, so $n = 5.0 / 342.0 = 1.46 \times 10^{-2}\,\mathrm{mol}$; $N = 1.46 \times 10^{-2} \times 6.02 \times 10^{23} = 8.8 \times 10^{21}$ [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule). Each has 12 carbon [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom): $12 \times 8.8 \times 10^{21} = 1.1 \times 10^{23}$ carbon [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom).

**Exercise 25.11 ★★.**

Which contains the larger amount of [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule): $1.0\,\mathrm{kg}$ of water $\ce{H2O}$ or $1.0\,\mathrm{kg}$ of ethanol $\ce{C2H6O}$? By what factor?

**Solution of Exercise 25.11.**

Water: $1000 / 18.0 = 55.6\,\mathrm{mol}$. Ethanol: $M = 2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0\,\mathrm{g}/\mathrm{mol}$, so $1000 / 46.0 = 21.7\,\mathrm{mol}$. Water has the larger amount, by a factor $55.6 / 21.7 \approx 2.6$, which is simply $46.0 / 18.0$.

**Exercise 25.12 ★★★.**

A gold ring has a mass of $5.0\,\mathrm{g}$; three quarters of its mass is gold, the rest copper. How many [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of gold does it contain? How many [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of copper? Which [metal](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal) has more [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) in the ring?

**Solution of Exercise 25.12.**

Gold: $3.75\,\mathrm{g}$, $n = 3.75 / 197.0 = 1.90 \times 10^{-2}\,\mathrm{mol}$, that is $1.15 \times 10^{22}$ [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom). Copper: $1.25\,\mathrm{g}$, $n = 1.25 / 63.5 = 1.97 \times 10^{-2}\,\mathrm{mol}$, that is $1.19 \times 10^{22}$ [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom). Surprisingly, the ring holds slightly more copper [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) than gold [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom): a gold [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) is about three times heavier than a copper [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom).

**Exercise 25.13 ★★★.**

A classroom measures $8.0\,\mathrm{m}$ by $6.0\,\mathrm{m}$ by $3.0\,\mathrm{m}$, and the [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) in it is at $20\,{}^{\circ}\mathrm{C}$.

1. What amount of gas does it contain ( $1\,\mathrm{m}^{3} =  1000\,\mathrm{L}$ )? How many [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) ?
2. Taking [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) as 78 [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of dinitrogen, 21 of dioxygen and 1 of argon ( $40.0\,\mathrm{g}/\mathrm{mol}$ ) in every hundred, compute the mean [molar mass](#def-g10-the-mole-molar-mass) of [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) .
3. What is the mass of the [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) in the room?

**Solution of Exercise 25.13.**

1. $V = 8.0 \times 6.0 \times 3.0 = 144\,\mathrm{m}^{3} =  1.44 \times 10^{5}\,\mathrm{L}$ ; $n = 1.44 \times 10^{5} / 24.1 = 5.98 \times 10^{3}\,\mathrm{mol}$ ; $N = 5.98 \times 10^{3} \times 6.02 \times 10^{23} = 3.6 \times 10^{27}$ [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) .
2. $M = (78 \times 28.0 + 21 \times 32.0 + 1 \times 40.0) / 100 =  2896 / 100 = 29.0\,\mathrm{g}/\mathrm{mol}$ .
3. $m = 5.98 \times 10^{3} \times 29.0 = 1.73 \times 10^{5}\,\mathrm{g}$ , about $173\,\mathrm{kg}$ : the [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) of a classroom weighs as much as two or three adults.

**Exercise 25.14 ★★★.**

A sphere of pure silicon has a mass of exactly $1.000\,\mathrm{kg}$.

1. How many silicon [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) does it contain?
2. Deduce the mass of one silicon [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) .
3. Before 2019 the [Avogadro constant](#def-g10-the-mole-avogadro) was not fixed but measured. Explain how an independent count of the [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of such a sphere, whose amount is known from its mass, would give a value of it.

**Solution of Exercise 25.14.**

1. $n = 1000 / 28.1 = 35.6\,\mathrm{mol}$ , so $N = 35.6 \times 6.02 \times 10^{23} = 2.14 \times 10^{25}$ [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) .
2. $m = 1.000\,\mathrm{kg} / 2.14 \times 10^{25} = 4.67 \times 10^{-26}\,\mathrm{kg}$ .
3. The mass of the sphere and the [molar mass](#def-g10-the-mole-molar-mass) give its amount, $n = m / M$ . If the [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) $N$ are counted independently (from the volume of the sphere and the spacing of the [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) in the crystal), the ratio $N / n$ is the [Avogadro constant](#def-g10-the-mole-avogadro) .

**Exercise 25.15 ★★★.**

Natural chlorine contains $75.8\,\%$ of chlorine-35 [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), of [molar mass](#def-g10-the-mole-molar-mass) very close to $35.0\,\mathrm{g}/\mathrm{mol}$, and $24.2\,\%$ of chlorine-37 [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), of [molar mass](#def-g10-the-mole-molar-mass) very close to $37.0\,\mathrm{g}/\mathrm{mol}$. Compute the [atomic molar mass](#def-g10-the-mole-molar-mass) of natural chlorine, and compare it with the value of the table.

**Solution of Exercise 25.15.**

$M = 0.758 \times 35.0 + 0.242 \times 37.0 = 26.53 + 8.95 =
35.5\,\mathrm{g}/\mathrm{mol}$: the value of the table.

## 25.7 Problem: Kelvin’s Glass of Water

**Problem 25.1.**

Weekend problem — pour a glass of water into the sea, wait for it to mix through all the oceans, and fill a glass again: how many of the first molecules come back?

A famous thought experiment, often attributed to the physicist Lord Kelvin, goes like this. Mark every [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) in a glass of water, pour the glass into the sea, and wait until the marked [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) have spread evenly through all the oceans of the world. Then dip the glass into the sea again, anywhere. Will it bring back any marked [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule)? The glass holds $250\,\mathrm{mL}$, and $1\,\mathrm{mL}$ of water has a mass of $1.0\,\mathrm{g}$. The oceans hold about $1.335 \times 10^{9}\,\mathrm{km}^{3}$ of water. Treat sea water as pure water for the counting.

**Part I — Water in a glass.**

1. Compute the [molar mass](#def-g10-the-mole-molar-mass) of water.
2. What is the mass of the water in the glass?
3. Compute the amount of water [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) in the glass.
4. What amount of hydrogen [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) does the glass contain? Of oxygen [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) ?

**Part II — [Molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) in the glass.**

5. How many water [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) does the glass hold?
6. Compute the mass of one water [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) , in grams.
7. Check the answers to 5 and 6 by multiplying them: what should the product be?
8. Counting one [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) per second, day and night, how many years would it take to count the [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of the glass? (A year lasts about $3.16 \times 10^{7}\,\mathrm{s}$ .)

**Part III — The ocean.**

9. Convert the volume of the oceans to cubic metres ( $1\,\mathrm{km}^{3} = 10^{9}\,\mathrm{m}^{3}$ ), then to litres.
10. Once the marked [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) have spread evenly, how many of them are there in each litre of ocean?
11. How many marked [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) does the second glass bring back?
12. Second method. Compute the amount of water in the oceans, taking $1.0\,\mathrm{kg}$ for the mass of one litre.
13. What fraction of all the ocean’s water [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) are marked?
14. Multiply this fraction by the number of [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of the second glass, and compare with the answer to question 11.

**Part IV — What it means.**

15. How many glasses of $250\,\mathrm{mL}$ could be filled with the water of the oceans?
16. Compare this number with the number of [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of one glass. Explain in one sentence why the second glass brings back so many marked [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) .
17. What volume of sea water must be drawn, on average, to bring back a single marked [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) ? Compare with a drop of $0.05\,\mathrm{mL}$ .
18. The answer to question 11 has many digits. Why should it be given only as an order of magnitude? Give two reasons.
19. State the final answer: about how many [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of the first glass come back in the second?

**Solution of Problem 25.1.**

**1.** $M(\ce{H2O}) = 2 \times 1.0 + 16.0 = 18.0\,\mathrm{g}/\mathrm{mol}$.

**2.** $m = 250 \times 1.0\,\mathrm{g} = 250\,\mathrm{g}$.

**3.** $n = 250 / 18.0 = 13.9\,\mathrm{mol}$.

**4.** Each [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) has two hydrogen [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) and one oxygen [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom): $27.8\,\mathrm{mol}$ of hydrogen [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), $13.9\,\mathrm{mol}$ of oxygen [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom).

**5.** $N = 13.9 \times 6.02 \times 10^{23} = 8.36 \times 10^{24}$ [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule).

**6.** $m = 18.0\,\mathrm{g}/\mathrm{mol} / 6.02 \times 10^{23}\,\mathrm{mol}^{-1} =
2.99 \times 10^{-23}\,\mathrm{g}$.

**7.** $8.36 \times 10^{24} \times 2.99 \times 10^{-23}\,\mathrm{g} = 250\,\mathrm{g}$, the mass of the water in the glass, as it should be.

**8.** $8.36 \times 10^{24} / 3.16 \times 10^{7} = 2.6 \times 10^{17}$ years: a number of years with seventeen zeros, far beyond any possible count.

**9.** $1.335 \times 10^{9}\,\mathrm{km}^{3} \times 10^{9} = 1.335 \times 10^{18}\,\mathrm{m}^{3}$, and with $1\,\mathrm{m}^{3} = 1000\,\mathrm{L}$, $1.335 \times 10^{21}\,\mathrm{L}$.

**10.** $8.36 \times 10^{24} / 1.335 \times 10^{21} = 6.26 \times 10^{3}$ marked [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) per litre.

**11.** $6.26 \times 10^{3} \times 0.250 = 1.57 \times 10^{3}$ marked [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) in the second glass.

**12.** Mass $1.335 \times 10^{21} \times 1.0\,\mathrm{kg} =
1.335 \times 10^{21}\,\mathrm{kg} = 1.335 \times 10^{24}\,\mathrm{g}$; $n = 1.335 \times 10^{24} / 18.0 =
7.42 \times 10^{22}\,\mathrm{mol}$.

**13.** $13.9 / 7.42 \times 10^{22} = 1.87 \times 10^{-22}$.

**14.** $1.87 \times 10^{-22} \times 8.36 \times 10^{24} = 1.57 \times 10^{3}$: the same answer as question 11, as it must be.

**15.** $1.335 \times 10^{21} / 0.250 = 5.34 \times 10^{21}$ glasses.

**16.** $8.36 \times 10^{24} / 5.34 \times 10^{21} \approx 1570$. A glass holds about 1600 times more [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) than the oceans hold glasses of water; spread evenly, the [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of one glass leave about 1600 in every glassful of ocean.

**17.** $1 / 6.26 \times 10^{3} = 1.6 \times 10^{-4}\,\mathrm{L} = 0.16\,\mathrm{mL}$, about three drops.

**18.** The ocean volume is an estimate known to three or four figures at best, and it was rounded; sea water is not pure water (it holds salts); the glass is not exactly $250\,\mathrm{mL}$; and the mixing through all the oceans would never be perfectly even. Only the order of magnitude can be trusted.

**19.** About 1600 [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of the first glass come back in the second.