A sample of calcium carbonate $\left(CaC{O}_3\right)$ has the following percentage composition : Ca = 40%; C = 12%; O = 48%. If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate obtained from another source will be 

Law of  constant proportions states that “ A sample of pure compound always contains same elements combined together in the same proportions by mass ”

Given  100g of ${\mathrm{CaCO}}_3$  contains 40g of  ‘Ca' or

$\frac{100}{25}=$4g of Calcium carbonate obtained from other source contains $\frac{40}{25}=1.6$g of Calcium