Question

A $5.5 \cdot g$ mass of a chromium chloride compound contains $1.82 \cdot g$ of metal. What is its empirical formula?

Answer 1

Hint: Empirical formula is the chemical formula which contains the simplest positive integer ratio of atoms present in the compound. The empirical formula can be calculated by using the mass of the component of the compound and mass of the whole compound. Chromium chloride is a metal chloride containing chromium as a metal component.

Complete answer:
We simply find the empirical formula for the chromium halide using the given grade. Because it is a binary compound, that is, there are only 2 components, so even if we only give the mass of the metal, we can easily calculate the mass of the metal and chloride in the sample. In each case, we divide the mass of each contribution by its atomic mass:

Now, we need to calculate moles of metal, molecular mass of chromium is $52.00 \cdot g \cdot mol$ , so the calculation can be given as $ = \dfrac{{1.82 \cdot g}}{{52.00 \cdot g \cdot mo{l^{ - 1}}}} = 0.0350 \cdot mol$
Now, the moles of chlorine present will be - $\dfrac{{5.5 \cdot g - 1.82 \cdot g}}{{35.45 \cdot g \cdot mo{l^{ - 1}}}} = 0.105 \cdot mol$
Now, the ratio can be calculated by dividing with smallest molar quantity, which is that of the metal:
Chromium = $\dfrac{{0.0350 \cdot mol}}{{0.0350 \cdot mol}} = 1$
Chloride = $\dfrac{{0.105 \cdot mol}}{{0.0350 \cdot mol}} = 3$
Thus, using the ratios calculated, the empirical formula can be derived as $CrC{l_3}$ . Hence, the finalised empirical formula will be $CrC{l_3}$ .

Note:
When we use oxidation numbers to distinguish compounds, we use Roman numerals instead of Arabic numerals.
Cr(iii) chloride = chromic chloride
Cr(ii) chloride = chromic chloride