$A$ sample of calcium carbonate has the following percentage composition: $Ca = 40 \%$,$C = 12 \%$,and $O = 48 \%$. According to the law of definite proportion,the weight of calcium in $4 \ g$ of a sample of calcium carbonate from another source will be (atomic weights: $Ca = 40, C = 12, O = 16$).
- A$1.6 \times 10^{-2} \ g$
- B$1.6 \ g$
- C$0.1 \ g$
- D$0.2 \ g$
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