# Chapter 4 - Exercises - Page 190: 107

The limiting reactant is $C_7H_6O_3$ The theoretical yield is equal to 1.63 g The percent yield is equal to 74.8 %

#### Work Step by Step

1. Calculate the mass of $C_4H_6O_3$: $$3.00 \space mL \times \frac{1.08 \space g}{1 \space mL} = 3.24 \space g$$ 2. Find the limiting reactant: - Calculate or find the molar mass for $C_4H_6O_3$: $C_4H_6O_3$ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 4 )+ ( 16.00 $\times$ 3 )= 102.09 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$3.24 \space g \times \frac{1 \space mole}{ 102.09 \space g} = 0.0317 \space mole$$ - Calculate or find the molar mass for $C_7H_6O_3$: $C_7H_6O_3$ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 7 )+ ( 16.00 $\times$ 3 )= 138.12 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$1.25 \space g \times \frac{1 \space mole}{ 138.12 \space g} = 0.00905 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.0317 \space mole \space C_4H_6O_3 \times \frac{ 1 \space mole \ C_9H_8O_4 }{ 1 \space mole \space C_4H_6O_3 } = 0.0317 \space mole \space C_9H_8O_4$$ $$0.00905 \space mole \space C_7H_6O_3 \times \frac{ 1 \space mole \ C_9H_8O_4 }{ 1 \space mole \space C_7H_6O_3 } = 0.00905 \space mole \space C_9H_8O_4$$ Since the reaction of $C_7H_6O_3$ produces less $C_9H_8O_4$ for these quantities, it is the limiting reactant. 3. Calculate the theoretical and percent yield: - Calculate or find the molar mass for $C_9H_8O_4$: $C_9H_8O_4$ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 9 )+ ( 16.00 $\times$ 4 )= 180.15 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.00905 \space mole \times \frac{ 180.15 \space g}{1 \space mole} = 1.63 \space g$$ $$Percent \space yield = \frac{ 1.22 }{ 1.63 } \times 100\% = 74.8 \%$$

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