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  4. 0.14 g of an element on combustion gives 0.28 g of its oxide. What is that element?

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Q. 0.14 g of an element on combustion gives 0.28 g of its oxide. What is that element?

EAMCETEAMCET 2010

Solution:

Eq. Weight of element $ \text{= }\frac{\text{Weight}\,\text{of}\,\text{element}}{\text{Weiight}\,\text{of}\,\text{oxygen}}\text{ }\!\!\times\!\!\text{ 8} $ $ =\frac{0.14}{0.28-0.14}\times 8=8{{g}^{g-equiv}} $ Thus, the element should be suphur as its eq. Weight can be 8. $ \underset{\begin{smallmatrix} \text{1}\,\text{mol} \\ \text{32}\,\text{g} \end{smallmatrix}}{\mathop{\text{S}}}\,+{{O}_{2}}\xrightarrow{{}}\underset{\begin{smallmatrix} \text{1}\,\text{mol} \\ \text{32 + (2}\,\times \,\,16\text{)}\,\,\text{=}\,\,\text{64}\,\text{g} \end{smallmatrix}}{\mathop{S{{O}_{2}}}}\, $ $ \because $ 32 g sulphur, on combustion, gives oxide $ =64\text{ }g $ $ \because $ 0.14 g sulphur, on combustion, will give oxide $ =\frac{64\times 0.14}{32}=0.28g $ Hence, it is confirmed that the element is sulphur.