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SAT II Chemistry : Avogadro's Number

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Example Questions

Example Question #1 : Avogadro's Number

How many nitrate ions are in one mole of $Ca(NO_{3})_{2}$ $\displaystyle Ca(NO_{3})_{2}$?

Possible Answers:

$3.0\cdot10^{23}$ $\displaystyle 3.0\cdot10^{23}$

$1.2\cdot10^{24}$ $\displaystyle 1.2\cdot10^{24}$

$2.4\cdot10^{24}$ $\displaystyle 2.4\cdot10^{24}$

$6.0\cdot10^{23}$ $\displaystyle 6.0\cdot10^{23}$

$1.8\cdot10^{24}$ $\displaystyle 1.8\cdot10^{24}$

Correct answer:

$1.2\cdot10^{24}$ $\displaystyle 1.2\cdot10^{24}$

Explanation:

Note that there are two nitrate ( $NO_{3}$ $\displaystyle NO_{3}$) ions for every formula unit of calcium nitrate. This means that in one mole of calcium nitrate, there are two moles of nitrate ions. To go from moles to ions, multiply 2 by Avogadro's number.

$2\cdot6.0\cdot10^{23} = 1.2\cdot10^{24}$ $\displaystyle 2\cdot6.0\cdot10^{23} = 1.2\cdot10^{24}$

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Example Question #2 : Stoichiometry

How many hydrogen atoms are in one mole of $(NH_{4})_{2}S$ $\displaystyle (NH_{4})_{2}S$?

Possible Answers:

$1.2\cdot10^{24}$ $\displaystyle 1.2\cdot10^{24}$

$6.0\cdot10^{23}$ $\displaystyle 6.0\cdot10^{23}$

$4.8\cdot10^{24}$ $\displaystyle 4.8\cdot10^{24}$

$2.4\cdot10^{24}$ $\displaystyle 2.4\cdot10^{24}$

$3.6\cdot10^{24}$ $\displaystyle 3.6\cdot10^{24}$

Correct answer:

$4.8\cdot10^{24}$ $\displaystyle 4.8\cdot10^{24}$

Explanation:

In one molecule of $(NH_{4})_{2}S$ $\displaystyle (NH_{4})_{2}S$, there are 2 ammonium ions, each of which has 4 hydrogen atoms for a total of 8 hydrogen atoms. Thus, to find the number of hydrogen atoms in one mole of $(NH_{4})_{2}S$ $\displaystyle (NH_{4})_{2}S$, one must multiply Avogadro's number by 8.

$8(6.0\cdot10^{23})=4.8\cdot10^{24}$ $\displaystyle 8(6.0\cdot10^{23})=4.8\cdot10^{24}$

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Example Question #3 : Stoichiometry

Calculate the number of aluminum ions in $4.5\times10^{-2}$ $\displaystyle 4.5\times10^{-2}$ moles of $\text{Al}_2(\text{Cr}_2\text{O}_7)_3$ $\displaystyle \text{Al}_2(\text{Cr}_2\text{O}_7)_3$_.

Possible Answers:

$6.6\times10^{22}\, \text{Al}^{3+}\,\text{ions}$ $\displaystyle 6.6\times10^{22}\, \text{Al}^{3+}\,\text{ions}$

$1.8\times10^{23}\, \text{Al}^{3+}\,\text{ions}$ $\displaystyle 1.8\times10^{23}\, \text{Al}^{3+}\,\text{ions}$

$3.3\times10^{21}\, \text{Al}^{3+}\,\text{ions}$ $\displaystyle 3.3\times10^{21}\, \text{Al}^{3+}\,\text{ions}$

$5.4\times10^{22}\,\text{Al}^{3+}\,\text{ions}$ $\displaystyle 5.4\times10^{22}\,\text{Al}^{3+}\,\text{ions}$

$4.8\times10^{22}\, \text{Al}^{3+}\,\text{ions}$ $\displaystyle 4.8\times10^{22}\, \text{Al}^{3+}\,\text{ions}$

Correct answer:

$5.4\times10^{22}\,\text{Al}^{3+}\,\text{ions}$ $\displaystyle 5.4\times10^{22}\,\text{Al}^{3+}\,\text{ions}$

Explanation:

In order calculate the number of aluminum ions, we must first find the number of aluminum ions in the entire compound. In this case, there are two molecules of $\text{Al}^{3+}$ $\displaystyle \text{Al}^{3+}$ for every molecule of $\text{Al}_2(\text{Cr}_2\text{O}_7)_3$ $\displaystyle \text{Al}_2(\text{Cr}_2\text{O}_7)_3$. After understanding this, we can use Avogadro's constant to determine the number of atoms (more specifically ions) of aluminum. See equation below for specific calculations.

$4.5\times10^{-2}\,\text{mol Al}_2(\text{Cr}_2\text{O}_7)_3\times\frac{2\;\text{mol Al}^{3+}}{1\,\text{mol Al}_2(\text{Cr}_2\text{O}_7)_3}$ $\displaystyle 4.5\times10^{-2}\,\text{mol Al}_2(\text{Cr}_2\text{O}_7)_3\times\frac{2\;\text{mol Al}^{3+}}{1\,\text{mol Al}_2(\text{Cr}_2\text{O}_7)_3}$ $\times\frac{6.02\times10^{23}\; \text{Al}^{3+}\,\text{ions}}{1\;\text{mol Al}^{3+}}=5.4\times10^{22}\; \text{Al}^{3+}\,\text{ions}$ $\displaystyle \times\frac{6.02\times10^{23}\; \text{Al}^{3+}\,\text{ions}}{1\;\text{mol Al}^{3+}}=5.4\times10^{22}\; \text{Al}^{3+}\,\text{ions}$

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Example Question #2 : Avogadro's Number

How many atoms of chloride are in 0.2550 g of aluminum chloride, $\text{AlCl}_3$ $\displaystyle \text{AlCl}_3$?

Possible Answers:

$4.125\times10^{21}\text{ atoms Cl}$ $\displaystyle 4.125\times10^{21}\text{ atoms Cl}$

$3.652\times10^{21}\text{ atoms Cl}$ $\displaystyle 3.652\times10^{21}\text{ atoms Cl}$

$3.454\times10^{21}\text{ atoms Cl}$ $\displaystyle 3.454\times10^{21}\text{ atoms Cl}$

$4.332\times10^{21}\text{ atoms Cl}$ $\displaystyle 4.332\times10^{21}\text{ atoms Cl}$

$3.823\times10^{21}\text{ atoms Cl}$ $\displaystyle 3.823\times10^{21}\text{ atoms Cl}$

Correct answer:

$3.454\times10^{21}\text{ atoms Cl}$ $\displaystyle 3.454\times10^{21}\text{ atoms Cl}$

Explanation:

The formula for aluminum chloride is $\text{AlCl}_3$ $\displaystyle \text{AlCl}_3$. The molar mass of $\text{Al}$ $\displaystyle \text{Al}$ is 26.98 g and the molar mass of $\text{Cl}$ $\displaystyle \text{Cl}$ is 35.45 g. There are three $\text{Cl}$ $\displaystyle \text{Cl}$ atoms in $\text{AlCl}_3$ $\displaystyle \text{AlCl}_3$. To calculate the molecular mass of $\text{AlCl}_3$ $\displaystyle \text{AlCl}_3$, we need to find the sum of the mass of one aluminum atom and three chlorine atoms:

$1(26.98\text{ amu Al})+3(35.45\text{ amu Cl})=133.33\text{ amu}$ $\displaystyle 1(26.98\text{ amu Al})+3(35.45\text{ amu Cl})=133.33\text{ amu}$

The total molecular weight is 130.33 g. Starting with the grams of $\text{AlCl}_3$ $\displaystyle \text{AlCl}_3$, we convert to the amount of moles $\text{AlCl}_3$ $\displaystyle \text{AlCl}_3$ using the molecular weight value we just calculated. Then we calculate the number of moles of chloride ion in $\text{AlCl}_3$ $\displaystyle \text{AlCl}_3$. using the 3:1 ratio. Finally, we calculate the number of atoms by using Avogadro's number. See calculations below:

$0.2550\text{ g AlCl}_3 \times \frac{1\text{ mol AlCl}_3}{133.33\text{ g AlCl}_3} \times\frac{3\text{ mol Cl}}{1\text{ mol AlCl}_3}$ $\displaystyle 0.2550\text{ g AlCl}_3 \times \frac{1\text{ mol AlCl}_3}{133.33\text{ g AlCl}_3} \times\frac{3\text{ mol Cl}}{1\text{ mol AlCl}_3}$ $\times\frac{6.02\times10^{23}\text{ atoms Cl}}{1\text{ mol Cl}}\,= 3.454\times10^{21}\text{ atoms Cl}$ $\displaystyle \times\frac{6.02\times10^{23}\text{ atoms Cl}}{1\text{ mol Cl}}\,= 3.454\times10^{21}\text{ atoms Cl}$

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SAT II Chemistry : Avogadro's Number

Study concepts, example questions & explanations for SAT II Chemistry

All SAT II Chemistry Resources

Example Questions

Example Question #1 : Avogadro's Number

Example Question #2 : Stoichiometry

Example Question #3 : Stoichiometry

Example Question #2 : Avogadro's Number

All SAT II Chemistry Resources

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