Answer
a. $$73.9 \space g \space C_4H_6O_3 $$
b. $$1.30 \times 10^{2} \space g \space C_9H_8O_4 $$
Work Step by Step
a.
$ C_7H_6O_3 $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 7 )+ ( 16.00 $\times$ 3 )= 138.12 g/mol
$$ \frac{1 \space mole \space C_7H_6O_3 }{ 138.12 \space g \space C_7H_6O_3 } \space and \space \frac{ 138.12 \space g \space C_7H_6O_3 }{1 \space mole \space C_7H_6O_3 }$$
$ C_4H_6O_3 $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 4 )+ ( 16.00 $\times$ 3 )= 102.09 g/mol
$$ \frac{1 \space mole \space C_4H_6O_3 }{ 102.09 \space g \space C_4H_6O_3 } \space and \space \frac{ 102.09 \space g \space C_4H_6O_3 }{1 \space mole \space C_4H_6O_3 }$$
$$ 1.00 \times 10^{2} \space g \space C_7H_6O_3 \times \frac{1 \space mole \space C_7H_6O_3 }{ 138.12 \space g \space C_7H_6O_3 } \times \frac{ 1 \space mole \space C_4H_6O_3 }{ 1 \space mole \space C_7H_6O_3 } \times \frac{ 102.09 \space g \space C_4H_6O_3 }{1 \space mole \space C_4H_6O_3 } = 73.9 \space g \space C_4H_6O_3 $$
b.
$ C_9H_8O_4 $ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 9 )+ ( 16.00 $\times$ 4 )= 180.15 g/mol
$$ \frac{1 \space mole \space C_9H_8O_4 }{ 180.15 \space g \space C_9H_8O_4 } \space and \space \frac{ 180.15 \space g \space C_9H_8O_4 }{1 \space mole \space C_9H_8O_4 }$$
$$ 1.00 \times 10^{2} \space g \space C_7H_6O_3 \times \frac{1 \space mole \space C_7H_6O_3 }{ 138.12 \space g \space C_7H_6O_3 } \times \frac{ 1 \space mole \space C_9H_8O_4 }{ 1 \space mole \space C_7H_6O_3 } \times \frac{ 180.15 \space g \space C_9H_8O_4 }{1 \space mole \space C_9H_8O_4 } = 1.30 \times 10^{2} \space g \space C_9H_8O_4 $$
