Answer
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:
$$ 0.00905 \space mole \times \frac{ 180.15 \space g}{1 \space mole} = 1.63 \space g$$
$$Percent \space yield = \frac{ 1.22 }{ 1.63 } \times 100\% = 74.8 \% $$
