# 6.6: Counting Particles By Weighing

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The following relationships have previously been established:

• 1 mol Al = 26.98 g Al = 6.022 × 1023 atoms Al
• 1 mol C12H22O11 = 342.3 g C12H22O11 = 6.022 × 1023 molecules C12H22O11
• 1 mol NaCl = 58.44 g NaCl = 6.022 × 1023 formula units NaCl

These relationships may be used to count atoms, molecules, or formula units by weighing just like we counted pennies by weighing.

#### Counting Atoms, Molecules, or Formula Units By Weighing

In the case of aluminum, here are the possible conversions:

 $${\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\;\mathrm{mol}\;\mathrm{Al}\;\;}}{\color[rgb]{0.8, 0.0, 0.0}\;}{\color[rgb]{0.1, 0.1, 0.1}\rightleftarrows}{\color[rgb]{0.1, 0.1, 0.1}\;}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\;\;\mathrm g\;\mathrm{Al}\;\;\;\;}}$$ $${\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\;\mathrm{mol}\;\mathrm{Al}\;\;}}\rightleftarrows{\color[rgb]{0.5, 0.0, 0.0}\boxed{\;\;\mathrm{atoms}\;\mathrm{Al}\;\;}}$$ $${\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\;\mathrm g\;\mathrm{Al}\;\;\;}}\rightleftarrows{\color[rgb]{0.5, 0.0, 0.0}\boxed{\;\;\mathrm{atoms}\;\mathrm{Al}\;\;}}$$ $$\frac{26.98\;\mathrm g\;\mathrm{Al}}{1\;\mathrm{mol}\;\mathrm{Al}}\;\mathrm{or}\;\frac{1\;\mathrm{mol}\;\mathrm{Al}}{26.98\;\mathrm g\;\mathrm{Al}}$$ $$\frac{6.022\times10^{23}\;\mathrm{atoms}\;\mathrm{Al}}{1\;\mathrm{mol}\;\mathrm{Al}}\;\mathrm{or}\;\frac{1\;\mathrm{mol}\;\mathrm{Al}}{6.022\times10^{23}\;\mathrm{atoms}\;\mathrm{Al}}$$ $$\frac{6.022\times10^{23}\;\mathrm{atoms}\;\mathrm{Al}}{26.98\;\mathrm g\;\mathrm{Al}}\;\mathrm{or}\;\frac{26.98\;\mathrm g\;\mathrm{Al}}{6.022\times10^{23}\;\mathrm{atoms}\;\mathrm{Al}}$$

When converting from grams of aluminum to atoms of aluminum, a conversion that places atoms of aluminum in the numerator and grams of aluminum in the denominator would be used to enure the proper cancellation of units:

$${\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\;\;\mathrm g\;\mathrm{Al}\;\;\;\;}}\xrightarrow[{26.98\;\mathrm g\;\mathrm{Al}}]{6.022\times10^{23}\;\mathrm{atoms}\;\mathrm{Al}}{\color[rgb]{0.5, 0.0, 0.0}\boxed{\;\mathrm{atoms}\;\mathrm{Al}\;}}$$

How many atoms are required to make the average empty aluminum beverage can that has a mass of 14.9 g?

$$14.9\;\cancel{\mathrm g\;\mathrm{Al}}\;\times\dfrac{6.022\times10^{23}\mathrm{atoms}\;\mathrm{Al}}{26.98\;\cancel{\mathrm g\;\mathrm{Al}}}=\boxed{3.33\times10^{23}\;\mathrm{atoms}\;\mathrm{Al}}$$

This calculation shows how it is possible to to count atoms by weighing using the definitions for molar mass and Avogadro's number.

✅ Example $$\PageIndex{1}$$

If a glass contains 9.64×1024 molecules H2O, how many moles of water are in the glass?

Solution

Steps for Problem Solving

Identify the "given" information and what the problem is asking you to "find."

Given: 9.64×1024 molecules H2O

Find: mol H2O

List known relationship(s).

1 mol H2O = 18.02 g H2O = 6.022 × 1023 molecules H2O

Prepare a concept map and use the proper conversion factor.

$${\color[rgb]{0.5, 0.0, 0.0}\boxed{\;\mathrm{molecules}\;{\mathrm H}_2\mathrm O\;}}\xrightarrow[{6.022\times10^{23}\;\mathrm{molecules}\;{\mathrm H}_2\mathrm O}]{1\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}{\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\;\;\;\;\mathrm{mol}\;{\mathrm H}_2\mathrm O\;\;\;\;\;}}$$

$$9.64\times10^{24}\;\cancel{\mathrm{molecules}\;{\mathrm H}_2\mathrm O}\times\dfrac{1\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}{6.022\times10^{23}\;\cancel{\mathrm{molecules}\;{\mathrm H}_2\mathrm O}}=\boxed{16.0\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}$$

The given number of water molecules was greater than Avogadro's number, so the number of moles of water molecules must be greater than 1 mole.

The answer is rounded to three significant figures since the fewest significant figures in the calculation is three.

✅ Example $$\PageIndex{2}$$

⅛ tsp of table salt (NaCl) has a mass of 0.69 g. How many formula units of NaCl are in this mass of salt?

Solution

Steps for Problem Solving

Identify the "given" information and what the problem is asking you to "find."

Given: 0.69 g NaCl

Find: formula units NaCl

List known relationship(s).

1 mol NaCl = 58.44 g NaCl = 6.022 × 1023 formula units NaCl

Prepare a concept map and use the proper conversion factor.

$${\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\;\;\mathrm g\;\mathrm{NaCl}\;\;\;\;}}\xrightarrow[{58.44\;\mathrm g\;\mathrm{NaCl}}]{6.022\times10^{23}\;\mathrm{formula}\;\mathrm{units}\;\mathrm{NaCl}}{\color[rgb]{0.5, 0.0, 0.0}\boxed{\mathrm{formula}\;\mathrm{units}\;\mathrm{NaCl}}}$$

$$0.69\;\cancel{\mathrm g\;\mathrm{NaCl}}\times\dfrac{6.022\times10^{23}\;\mathrm{formula}\;\mathrm{units}\;\mathrm{NaCl}}{58.44\;\cancel{\mathrm g\;\mathrm{NaCl}}}=\boxed{7.1\times10^{21}\;\mathrm{formula}\;\mathrm{units}\;\mathrm{NaCl}}$$

There should be a huge number of formula units of NaCl. However, 0.69 g NaCl is quite a bit less than the mass of one mole NaCl (58.44 g), so there should be less than Avogadro's number of formula units.

The answer is rounded to two significant figures since the fewest significant figures in the calculation is two.

✏️ Exercise $$\PageIndex{2}$$

1. What is the mass of one molecule of water, expressed in units of grams?  Provide your answer to 3 significant figures.
2. Which of the following amounts contains the fewest atoms?
12.01 g C     26.00 g Cr     40.08 g Ca     58.93 g Co     63.55 g Cu