Answer to Question #112247 in General Chemistry for torrey

Question

Calculate the percent mass of EACH element in isoamyl acetate: C7H14O2 *

Answer 1

Solution

M(C7H14O2) = 130.1849 g/mol

$w(X) = \frac{n \times (X)}{M(C7H14O2)}*100 \%$

$w(C) = \frac{7 \times 12.0107}{130.1849}*100 \% = 64.58 \%$

$w(H) = \frac{14 \times 1.0079}{130.1849}*100 \% = 10.84 \%$

$w(O) = \frac{2 \times 16.00}{130.1849}*100 \% = 24.58 \%$

Answer:

w(C) = 64.58 %

w(H) = 10.84 %

w(O) = 24.58 %