Question

Impure sample contains one mole of glucose $\left(C_6{H}_{12}{O}_6\right)$ and impurity. In this simple oxygen is present $50$% by mass, then what is the mas percentage of hydrogen in sample?

(Assume impurity does not contain $C,H$ and $O$)

Answer 1

% Oxygen= $\frac{\text{mass of oxygen}}{\text{total mass of sample}}\times 100$

$\therefore 50=\frac{6\times 18}{m}\times 100$

$\therefore m=192g$

$\therefore$ % hydrogen= $\frac{12\times 1}{192}\times 100$

$=6.25$%