Answer:
The empirical formula of the compound = [tex]K_1C_1O_3[/tex]
The molecular formula of the compound = [tex]K_2C_2O_6[/tex]
Explanation:
Mass of potassium = 7.89 g
Molar mass of potassium = 39 g/mol
Moles of potassium = [tex]\frac{7.89 g}{39 g/mol}=0.2023 mol[/tex]
Mass of carbon = 2.42 g
Molar mass of carbon = 12 g/mol
Moles of carbon = [tex]\frac{2.42 g}{12 g/mol}=0.2017 mol[/tex]
Mass of oxygen = 9.69 g
Molar mass of oxygen = 16 g/mol
Moles of oxygen = [tex]\frac{9.69 g}{16g/mol}=0.6056 mol[/tex]
For the empirical mass of the compound, divide the lowest moles of an element with all the moles of the elements:
Potassium : [tex]\frac{0.2023 mol}{0.2017 mol}=1[/tex]
Carbon : [tex]\frac{0.2017 mol}{0.2017 mol}=1
Oxygen : [tex]\frac{0.6056 mol}{0.2017 mol}=3[/tex]
The empirical formula of the compound = [tex]K_1C_1O_3[/tex]
Empirical mass of the compound= (39 g/mol+12 g/mol+3 × 16 g/mol) = 99 g/mol
Molar mass of the compound = 198.2 g/mol
Let the molecular formula of the compound be : [tex]K_{1n}C_{1n}O_{3n}[/tex]
[tex]n=\frac{\text{Molar mass}}{\text{Empirical mass}}[/tex]
[tex]n=\frac{198.22 g/mol}{99 g/mol}=2[/tex]
So, molecular mas of the compound :
[tex]K_{1\times 2}C_{1\times 2}O_{3\times 2}=K_2C_2O_6[/tex]
