Question

A sample of pure calcium fluoride with a mass of 15.0 g contains 7.70 g of calcium. how much calcium is contained in 45.0 g of calcium fluoride?

Answer 1

In a sample of pure calcium fluoride of mass 15.0 g, 7.70 g of calcium is present. First convert the mass into number of moles as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Molar mass of Ca is 40 g/mol, putting the values,

$n=\frac{7.70 g}{40 g/mol}=0.1925 mol$

Similarly, molar mass of $CaF_{2}$ is 78.07 g/mol thus, number of moles will be:

$n=\frac{15.0 g}{78.07 g/mol}=0.1921 mol$.

Thus, 0.1921 mol of $CaF_{2}$ have 0.1925 mol of Ca, or 1 mole of $CaF_{2}$ will have approximately 1 mole of Ca.

Now, mass of Ca needs to be calculated in 45.0 g of $CaF_{2}$. Converting mass into number of moles first,

$n=\frac{45.0 g}{78.07 g/mol}=0.5764 mol$

Thus, number of moles of Ca will also be 0.5764 mol, converting number of moles into mass,

$m=n\times M=0.5764 mol\times 40 g/mol=23.06 g$

Therefore, mass of Ca will be 23.06 g.