Question

Item 9 A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, the percent of CaCO3 in the sample is

Answer 1

Answer:

The percent of CaCO₃ in the sample is 75%.

Explanation:

We can find the percent of CaCO₃ in the sample knowing that 30 g of Ca are present in 100 g of CaCO₃.

First, we need to find the number of moles of Ca, and since the ratio of Ca in CaCO₃ is 1:1 we can find the percent of CaCO₃ in the sample as follows:

$mol CaCO_{3} = mol Ca = \frac{30 g}{40 g/ 1 mol} = 0.75 mol$

Now, the mass of CaCO₃ is:

$m CaCO_{3} = 0.75 moles*100 g/mol = 75 g$

Finally, the percent of CaCO₃ in 100 g of the sample is:                                        

$\% CaCO_{3} = \frac{75 g}{100 g}*100 = 75 \%$                  

   

Therefore, the percent of CaCO₃ in the sample is 75%.

I hope it helps you!