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# A sample of calcium carbonate ($CaC{O}_{3}$) has the following percentage composition:Ca = 40%; C = 12%; O = 48%.If the law constant proportions is true, then the weight of calcium in 4 g of a sample calcium carbonate from another source will be:A. 0.016gB. 0.16gC. 1.6gD. 16 g

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Hint: Law of constant proportions is a law that implies that the sample of a compound will contain the same elements that are present in the same ratio by mass as that of any other sample of the same compound.

Complete answer:
The law of constant proportions states that the chemical compounds are made up of elements that are present in a fixed ratio by mass. This means that a pure sample of a compound, no matter what the source is, will always contain the same elements that are present in the same ratio by mass. Let us now use this law to find the weight of the calcium in 4 g of calcium carbonate.
The percentage compositions of the elements of calcium carbonate are given as: Ca = 40%; C = 12%; O = 48%. As it is also given that the law of constant proportion holds true, then this means that even the sample of 4g of calcium carbonate will have the percentage composition of the constituent elements as: Ca = 40%; C = 12%; O = 48%.
Therefore, this means that 100g of $CaC{O}_{3}$, 40 g of Ca is present.
Thus, in a 4g sample of $CaC{O}_{3}$, let x g of Ca be present.

This means, $\dfrac {x}{4} = \dfrac {40}{100}$

$\implies x = \dfrac {40 \times x}{100}$
$\implies x = 1.6g$

Therefore, the weight of calcium in 4 g of a sample calcium carbonate from another source is 1.6g.

Hence, the correct answer is option (C).

Note: The law of constant proportions does not hold true in all the cases. There are some exceptions too. Some non-stoichiometric compounds have different element composition in different samples. These compounds rather obey the law of multiple proportions.