Question

data sheet
1 mass of crucible + cover 31.885 g
2 mass of crucible + cover + mg 32.072 g
3 mass of mg 0.187 g
4 mass of crucible + cover + oxide product 32.194 g
5 mass of oxide product 0.309 g
32.194 - 31.885 = 0.309
6 mass of oxygen in oxide product 0.122 g
0.309 - 0.187 = 0.122
7 mgo(s)
empirical formula of oxide product (show calculations to receive credit)
0.187g mg × \frac{1mol mg}{24.31g mg}=\frac{0.00769 mol mg}{0.00763}=1
0.122g o × \frac{1mol o}{16g o}=\frac{0.00763 mol o}{0.00763}=1

Explanation:

the molar - mass of Mg ($M_{Mg}=24.31\ g/mol$) to convert the mass of Mg to moles. The formula is $n=\frac{m}{M}$. So, $n_{Mg}=\frac{0.187\ g}{24.31\ g/mol}= 0.00769\ mol$.

Step2: Calculate moles of O

We use the molar - mass of O ($M_O = 16\ g/mol$) to convert the mass of O to moles. The formula is $n=\frac{m}{M}$. So, $n_O=\frac{0.122\ g}{16\ g/mol}=0.00763\ mol$.

Step3: Determine the ratio of moles

We divide the number of moles of each element by the smaller number of moles to get the ratio of atoms in the empirical formula. $\frac{n_{Mg}}{n_{min}}=\frac{0.00769\ mol}{0.00763\ mol}\approx1$ and $\frac{n_O}{n_{min}}=\frac{0.00763\ mol}{0.00763\ mol} = 1$.

Answer:

The empirical formula of the oxide product is MgO.