# Question 8d3dd

Jun 10, 2017

1, Staring material $0.25 g$ compound

Whole Ca of the compound gives $0.16 g$ $C a C {O}_{3}$

Molar mass of $C a C {O}_{3} \to 100 g \text{/} m o l$

So $0.25 g$ compound contains $0.16 \times \frac{40}{100} g$ Ca

Hence %of Ca->0.16xx40/100xx100/0.25=25.60

2, Staring material $0.115 g$ compound

Whole S of the compound gives $0.344 g$ $B a S {O}_{4}$

Molar mass of $B a S {O}_{4} \to 233 g \text{/} m o l$

So $0.115 g$ compound contains $0.344 \times \frac{32}{233} g$ Sulphur

Hence %of S->0.344xx32/233xx100/0.115=40.87

3, Staring material $0.712 g$ compound

Whole N of the compound gives $0.155 g$ $N {H}_{3}$

Molar mass of $N {H}_{3} \to 17 g \text{/} m o l$

So $0.712 g$ compound contains $0.115 \times \frac{14}{17} g$ Nitrogen

Hence %of N->0.115xx14/17xx100/0.712=17.98

So %of C=(100-25.60+40.87+17.98)=15.55#

The ratio of number atoms of

$C a : C : N : S = \frac{25.6}{40} : \frac{15.55}{12} : \frac{17.98}{14} : \frac{40.87}{32}$

$\implies C a : C : N : S = 0.64 : 1.29 : 1.28 : 1.28$

$\implies C a : C : N : S = 1 : 2 : 2 : 2$

So empirical formula $C a {\left(C N S\right)}_{2}$
The empirical formula mass is $40 + 2 \left(12 + 14 + 32\right) = 156 g \text{/} m o l$
The given molar mass being same , the molecular formula is also same.