Question

(d) why will the crucible and its contents gain mass as a result of heating the mg in air?

1. a student determined the empirical formula of potassium oxide using the procedure of this experiment. she obtained the following data:

mass of crucible and cover, g: 28.288
mass of crucible, cover, and k, g: 28.709
mass of crucible, cover, and potassium oxide, g: 28.793
use these data to determine the empirical formula of potassium oxide. the molar mass of potassium (k) is 39.10 g/mol.

Explanation:

Step1: Calculate mass of K

Mass of \(K=m(\text{crucible, cover, and }K)-m(\text{crucible and cover})\)
\(m(K)=28.709 - 28.288=0.421\space g\)

Step2: Calculate moles of K

Moles of \(K,n(K)=\frac{m(K)}{M(K)}\), where \(M(K) = 39.10\space g/mol\)
\(n(K)=\frac{0.421}{39.10}\approx0.01077\space mol\)

Step3: Calculate mass of oxygen

Mass of oxygen \(m(O)=m(\text{crucible, cover, and potassium oxide})-m(\text{crucible, cover, and }K)\)
\(m(O)=28.793 - 28.709 = 0.084\space g\)

Step4: Calculate moles of O

Moles of \(O,n(O)=\frac{m(O)}{M(O)}\), where \(M(O)=16.00\space g/mol\)
\(n(O)=\frac{0.084}{16.00}=0.00525\space mol\)

Step5: Find mole ratio

Divide moles of each element by the smaller number of moles.
\(\frac{n(K)}{n(O)}=\frac{0.01077}{0.00525}\approx2\), \(\frac{n(O)}{n(O)} = 1\)

Answer:

The empirical formula of potassium oxide is \(K_{2}O\)