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AP Chemistry : Other Stoichiometric Calculations

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Example Questions

Example Question #21 : Stoichiometry

You have an unidentified colorless, odorless liquid in a thin cubic container. The container is 3.2 inches by 3.4 inches by 3.3 inches and the liquid fills the entire container. The mass of the liquid minus the mass of the container is 0.370 pounds. What is the liquid's density in grams per milliliter?

(1 inch = 2.54 cm and 1 kg = 2.2046 pounds)

Possible Answers:

$0.364\frac{g}{mL}$

$1.32\frac{g}{mL}$

$1.84\hspace{1 mm}\frac{g}{mL}$

$0.285\frac{g}{mL}$

$0.827\frac{g}{mL}$

Correct answer:

$0.285\frac{g}{mL}$

Explanation:

First we must determine the volume of the container:

$3.2\hspace{1 mm}in\times 3.4\hspace{1 mm} in\times 3.3\hspace{1 mm} in= 35.9\hspace{1 mm}in^3$

The density is $\frac{0.370\hspace{1 mm}lbs}{35.9\hspace{1 mm}in^3}$ .

Now we will convert this into grams per milliliter.

$\frac{0.370lbs}{35.9in^3}*(\frac{1in}{2.54cm})^3*\frac{1cm^3}{1mL}=6.29*10^{-4}\frac{lb}{mL}$

$6.29*10^{-4}\frac{lb}{mL}*\frac{1kg}{2.2046lb}*\frac{1000g}{1kg}=0.285\frac{g}{mL}$

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Example Question #22 : Stoichiometry

How much heat is released from the reaction of 18 moles of methane if each mole of methane liberates 5kJ of heat, but the reaction is only 60% efficient?

Possible Answers:

5kJ

90kJ

54kJ

60kJ

18kJ

Correct answer:

54kJ

Explanation:

Each mole of methane to react releases 5kJ of heat. Since we have 18 moles of methane, we expect a potential of 90kJ of heat to be released.

$18mol * \frac{5kJ}{mol} = 90kJ$

We also have to take into account that the process is only 60% efficient; therefore, we effectively only liberate 60% of the total expected amount.

$18mol * 0.60 * \frac{5kJ}{mol} = 54kJ$

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Example Question #3 : Other Stoichiometric Calculations

When $5.884\,g$ of NaCl are dissolved in $100\,mL$ of a $1.00\,M$ solution of $AgNO_{3}$

Possible Answers:

after filtration $170\,g$ of a solid are obtained

we obtain a clear solution

the approximate concentration of $NO{_{3}}^{-}$ in the resulting solution will be $1.00\,M$ and after filtration $17.0\,g$ of a solid are obtained

a precipitate of sodium nitrate will be obtained

Correct answer:

the approximate concentration of $NO{_{3}}^{-}$ in the resulting solution will be $1.00\,M$ and after filtration $17.0\,g$ of a solid are obtained

Explanation:

The net ionic equation that occurs is:

$Ag_{(aq)}^{+}\; +\;Cl _{(aq)}^{-}\; \rightarrow \; AgCl_{(s)}$

$NO_{3}^{-}$ and $Na^{+}$ are spectator ions. Hence, $NO_{3}^{-}$ concentration is not affected. The molecular mass of NaCl is 58.44 g , then 5.844 g of NaCl yields $0.1000 \: mol$ of $Cl^{-}$ . Since we have $0.100\: L$ of a $1.00 \: M$ solution of $AgNO_{3}$ , we have $0.100\: mol$ of $Ag^{+}$ . There is no limiting reactant. Hence $0.100\: mol$ of AgCl are formed. Being the molecular mass of $AgCl \: 169.9 \: g/mol$ , $17.0 g \: of \: AgCl$ are formed. Result should be given with tree significant figures.

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Example Question #1 : Other Stoichiometric Calculations

What is the mass of $8.888 * 10^{9} atoms \ Fe$ ?

Possible Answers:

$8.24 * 10^{-13} g\ Fe$

$2.989 * 10^{35} g\ Fe$

$2.64 * 10^{-16} g\ Fe$

$8.243 * 10^{-13} g\ Fe$

$2.99 * 10^{35} g\ Fe$

Correct answer:

$8.243 * 10^{-13} g\ Fe$

Explanation:

$8.888 * 10^{9} atoms \ Fe * \frac{1 mol Fe}{6.022 * 10^{23}atoms\ Fe} * \frac{55.85 g Fe}{1 mol Fe} = 8.243 * 10^{-13} g Fe$

Remember: the given measurement has 4 significant figures, so the answer must also have 4 significant figures.

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Example Question #2 : Other Stoichiometric Calculations

How many atoms are there in 17.0g of sulfur?

Possible Answers:

$3.2 * 10^{23} atoms\ S$

$3.28 * 10^{26} atoms\ S$

$3.3 * 10^{26} atoms\ S$

$3.19 * 10^{23} atoms\ S$

$8.81 * 10^{-25} atoms\ S$

Correct answer:

$3.19 * 10^{23} atoms\ S$

Explanation:

$17.0 g\ S * \frac{1 mol S}{32.06 g\ S} * \frac{6.022^{23} atoms \ S}{1 mol\ S} = 3.19 * 10^{23} atoms \ S$

Remember, there really isn't a way to go straight from grams to atoms, but, it's possible to change grams into moles (using the molar mass), and it's possible to change moles into atoms (using Avogadro's number).

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Example Question #31 : Stoichiometry

For the molecular formula, $CH_{3}CH_{2}Br$ , how many moles of bromine are in 20.0 grams of this molecule?

Possible Answers:

1.10mol

20.55mol

0.1835mol

8.1112mol

15.0mol

Correct answer:

0.1835mol

Explanation:

The first step is to determine the molecular weight of the molecule:

$MW=(atomic\ weight\ of\ C)+(atomic\ weight\ of\ H)+(atomic\ weight\ of\ Br)$

$MW_{CH_{3}CH_{2}Br}=2(12\frac{g}{mole})+5(1\frac{g}{mole})+(80\frac{g}{mole})=109 \frac{g}{mole}$

The molecular weight serves as a conversion factor to convert grams to moles as implied its units $\frac{g}{mole}$ . So, given that we are dealing with 20.0 grams, the moles of 20.0 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

$20.0 g * \frac{mole}{109\ grams} =0.1835\ moles\ CH_{3}CH_{2}Br$

Based on the molecular formula, for every mole of the compound $CH_{3}CH_{2}Br$ , there is 1 mole of bromine. Based on our calculations 20.0 grams of $CH_{3}CH_{2}Br$ is 0.1835 moles of the compound. Because there is a 1:1 molar ratio of bromine to every molecule, we have 0.1835 moles of bromine in 20 grams of our compound, $CH_{3}CH_{2}Br$ .

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Example Question #32 : Stoichiometry

For the molecular formula, $CH_{3}CH_{2}OH$ , how many grams of oxygen are in 50.0g of this molecule?

Possible Answers:

17.4g

16.2g

46.1g

12.4g

20.7g

Correct answer:

17.4g

Explanation:

The first step is to determine the molecular weight of the molecule ethanol, $CH_{3}CH_{2}OH$ :

$MW=\text{respective equivalents *[(atomic weight of C)+(atomic weight of H)+(atomic weight of O)]}$

$MW_{CH_{3}CH_{2}OH}=2\left(\frac{12g}{mol}\right)+6\left(\frac{1g}{mol}\right)+\left(\frac{16g}{mol}\right)=46 \frac{g}{mol}$

The molecular weight serves as a conversion factor to convert grams to moles as implied by its units $\frac{grams}{mole}$ . So, given that we are dealing with 50 grams of ethanol, the moles of 50 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

$50g * \frac{1mole}{46 g} =1.09 mol\ CH_{3}CH_{2}OH$

Based on the molecular formula, we can see that for every mole of the compound $CH_{3}CH_{2}OH$ , there is 1 mole of oxygen. Based on our calculations, 50 grams of $CH_{3}CH_{2}OH$ is 1.09 moles of the compound. Because there is a 1:1 molar ratio of oxygen to every ethanol molecule, that is, 1 mole of oxygen in every mole of our molecule, we have 1.09 moles of oxygen in 50 grams of our compound $CH_{3}CH_{2}OH$ . To convert 1.09 moles of oxygen to grams of oxygen, we perform the following equation based on the atomic weight of oxygen, which from the periodic table we know is $16\frac{g}{mol}$ :

$1.09 mol * \frac{16g}{mol}= 17.4 g\ oxygen$

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Example Question #6 : Other Stoichiometric Calculations

For the molecular formula, $C_{4}H_{9}O_{2}$ , how many grams of oxygen are in 45.0 grams of this molecule?

Possible Answers:

10.1g

25.4g

11.4g

55.0g

16.2g

Correct answer:

16.2g

Explanation:

The first step is to determine the molecular weight of the molecule:

$MW=\text{respective equivalents *[(atomic weight of C)+(atomic weight of H)+(atomic weight of O)]}$

$MW_{C_{4}H_{9}O_{2}}=4(12\frac{g}{mole})+9(1\frac{g}{mole})+2(16\frac{g}{mole})= 89 \frac{g}{mole}$

The molecular weight serves as a conversion factor to convert grams to moles as implied its units $\frac{g}{mole}$ . So, given that we are dealing with 45.0 grams, the moles of 45.0 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

$45.0g * \frac{mole}{89.0\ grams} =0.506\ moles C_{4}H_{9}O_{2}$

Based on the molecular formula, we can see that for every mole of the compound $C_{4}H_{9}O_{2}$ , there are 2 moles of oxygen. Based on our calculations 45.0 grams of $C_{4}H_{9}O_{2}$ is 0.506 moles of the compound. Because there is a 2:1 mole ratio of oxygen to every molecule, that is, 2 mole of oxygen in every mole of our molecule, we have double the moles of oxygen in our compound $C_{4}H_{9}O_{2}$ as calculated below:

$0.506\ moles_{C_{4}H_{9}O_{2}} * \frac{2\ moles\ of\ oxygen}{1\ mole_{C_{4}H_{9}O_{2}}}=1.011\ moles\ of\ oxygen$

To convert 1.011 moles of oxygen to grams of oxygen, we perform the following equation based on the atomic weight of oxygen, which is $16\frac{g}{mole}$ :

$1.012\ moles * \frac{16g}{mole}= 16.2 g\ oxygen$

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AP Chemistry : Other Stoichiometric Calculations

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #21 : Stoichiometry

Example Question #22 : Stoichiometry

Example Question #3 : Other Stoichiometric Calculations

Example Question #1 : Other Stoichiometric Calculations

Example Question #2 : Other Stoichiometric Calculations

Example Question #31 : Stoichiometry

Example Question #32 : Stoichiometry

Example Question #6 : Other Stoichiometric Calculations

All AP Chemistry Resources

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