A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

Question

Vladimir79

2 months ago

11

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

alculate the percent of CaCO3 in the sample.

Chemistry

1 answer:

Anarel [2.9K] · Answer 1 · 2026-02-17T03:07:11+03:00

Answer:

About 75%.

Explanation:

Refer to the atomic mass of Ca on a current periodic table:

Ca: 40.078.

A mole of Ca atoms is present in each mole of the CaCO₃ compound.

The molar mass of one mole of CaCO₃ corresponds to the mass of that compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms matches numerically with the relative atomic mass of this element: $\rm 40.078\; g$ .

Determine the mass ratio of Ca within a pure CaCO₃ sample:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Assume the mass of the sample is 100 g. This CaCO₃ sample is comprised of 30% Ca by mass. In that 100 grams, there would be $\rm 30 \% \times 100\; g = 30\; g$ Ca atoms. Assuming that there are no Ca impurities. Effectively, all of these Ca atoms are part of CaCO₃. Utilize the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

Thus, according to these assumptions, 100 grams of this sample would consist of 75 grams of CaCO₃. Consequently, the percentage mass of CaCO₃ in this sample would be:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .