Question

What is the percent composition of Chromium in Chromium (iii) Carbonate?

Answer 1

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the reaction between the ions Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

`Cr^(3+)+CO^(2-)_3\rightarrow Cr_2(CO_3)_3`

Step 2 - Finding the molar mass

The molar mass of Chromium (III) Carbonate can be found by multiplying the molar masses of each element by the number of times it appers in the formula and, finally, summing it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

`\begin{gathered} C\rightarrow3*12=36\text{ } \\ \\ O\rightarrow9*16=144 \\ \\ Cr\rightarrow2*52=104 \end{gathered}`

The molar mass will be thus:

`M=36+144+104=284\text{ g/mol}`

Step 3 - Finding the percent composition of Chromium

We saw in step 2 that the molar mass of Cr2(CO3)3 is 284 g/mol. From this mass, 104 g come from Cr. Therefore, we can set the following proportion:

`\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 104\text{ g/mol ---- x} \\ \\ x=(104*100)/(284)=(10400)/(284)=36.6\text{ \%} \end{gathered}`

The percent composition of Chromium is thus 36.6 %.