Stoichiometry — Hard Practice Quiz

A Chemistry cheat sheet for Stoichiometry — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.

Practice quiz

1. A sample of an unknown organic compound contains $1.204 \times 10^{24}$ atoms of carbon. If the compound is $40.0\%$ carbon by mass, and the atomic weight of carbon is $12.01$ g/mol, what is the total mass of the compound? Use Avogadro's number $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.

   • $60.05$ g
   • $120.1$ g
   • $240.2$ g
   • $30.03$ g

   Answer: $60.05$ g

2. In a reaction, $10.0$ g of reactant A (molar mass $50.0$ g/mol) reacts completely to form product B (molar mass $100.0$ g/mol) according to the balanced equation $A \rightarrow B$. If the reaction has a $75.0\%$ yield, how many molecules of B are actually produced? Use Avogadro's number $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.

   • $9.033 \times 10^{22}$ molecules
   • $1.204 \times 10^{23}$ molecules
   • $7.528 \times 10^{22}$ molecules
   • $1.506 \times 10^{23}$ molecules

   Answer: $9.033 \times 10^{22}$ molecules

3. Consider the reaction $2Fe + 3Cl_2 \rightarrow 2FeCl_3$. If $111.7$ g of Fe (atomic weight $55.845$ g/mol) reacts with excess $Cl_2$ (atomic weight $35.45$ g/mol) and the reaction has an $80.0\%$ yield, what mass of chlorine is present in the actual yield of $FeCl_3$?

   • $170.2$ g
   • $212.7$ g
   • $136.1$ g
   • $106.4$ g

   Answer: $170.2$ g

4. If the actual yield of a reaction is doubled, while the theoretical yield remains constant, how does the percent yield change? And if the number of atoms of a specific element in a compound is doubled, how does its mass percentage in that compound change, assuming the atomic weights of all elements and the number of atoms of other elements remain constant?

   • Percent yield doubles; Mass percentage increases but does not necessarily double.
   • Percent yield doubles; Mass percentage doubles.
   • Percent yield halves; Mass percentage increases but does not necessarily double.
   • Percent yield halves; Mass percentage doubles.

   Answer: Percent yield doubles; Mass percentage increases but does not necessarily double.

5. A compound has a molar mass of $180.0$ g/mol. It is found to contain $53.3\%$ oxygen by mass. How many oxygen atoms are present in $1.0$ mole of this compound? Use the atomic weight of oxygen as $16.00$ g/mol and Avogadro's number $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.

   • $3.611 \times 10^{24}$ atoms
   • $1.807 \times 10^{24}$ atoms
   • $6.022 \times 10^{23}$ atoms
   • $1.204 \times 10^{24}$ atoms

   Answer: $3.611 \times 10^{24}$ atoms

6. In the reaction $2H_2 + O_2 \rightarrow 2H_2O$, $4.0$ g of $H_2$ (molar mass $2.016$ g/mol) reacts with $32.0$ g of $O_2$ (molar mass $32.00$ g/mol). If $30.0$ g of $H_2O$ (molar mass $18.016$ g/mol) is collected, what is the percent yield of the reaction?

   • $83.9\%$
   • $93.7\%$
   • $75.0\%$
   • $100.0\%$

   Answer: $83.9\%$

7. A compound $X_2Y_3$ has a mass percentage of $X$ equal to $P\%$. If the atomic weight of $Y$ is $A_Y$, what is the atomic weight of $X$ ($A_X$) in terms of $P$ and $A_Y$?

   • $A_X = \frac{3 P A_Y}{200 - 2 P}$
   • $A_X = \frac{2 P A_Y}{100 - 3 P}$
   • $A_X = \frac{P A_Y}{100 - P}$
   • $A_X = \frac{3 P A_Y}{100 - P}$

   Answer: $A_X = \frac{3 P A_Y}{200 - 2 P}$

8. If a reaction has a $75.0\%$ yield and the theoretical yield of $C_6H_{12}O_6$ (molar mass $180.156$ g/mol, atomic weight of H is $1.008$ g/mol) is $100.0$ g, how many hydrogen atoms are present in the actual yield of the product? Use Avogadro's number $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.

   • $3.008 \times 10^{24}$ atoms
   • $4.011 \times 10^{24}$ atoms
   • $2.256 \times 10^{24}$ atoms
   • $3.760 \times 10^{24}$ atoms

   Answer: $3.008 \times 10^{24}$ atoms

9. A student performs a reaction and calculates the theoretical yield. They then measure the actual yield. If the student accidentally used an atomic weight for the limiting reactant that was $10\%$ higher than its true value, how would this error affect the calculated percent yield?

   • The calculated percent yield would be higher than the true percent yield.
   • The calculated percent yield would be lower than the true percent yield.
   • The calculated percent yield would be unaffected.
   • The effect on percent yield cannot be determined without knowing the actual values.

   Answer: The calculated percent yield would be higher than the true percent yield.

10. Consider a compound $M_xO_y$. If $1.0$ g of this compound contains $N$ atoms of metal $M$, and the mass percentage of metal $M$ in the compound is $P_M\%$, what is the atomic weight of $M$ ($A_M$) in terms of $P_M, N,$ and $N_A$ (Avogadro's number)?

    • $A_M = \frac{P_M \times N_A}{100 N}$
    • $A_M = \frac{100 N}{P_M \times N_A}$
    • $A_M = \frac{P_M N}{100 N_A}$
    • $A_M = \frac{100 N_A}{P_M N}$

    Answer: $A_M = \frac{P_M \times N_A}{100 N}$