Question

1. what is the value of x when the empirical formula is c₅hₓ and the molecular mass is 260.6 g/mol. a) 5 b) 0.02 c) 160 d) 10 e) none of the above 30) determine the empirical formula of a compound containing 83% potassium and 17% oxygen. a) ko b) k₂o₃ c) ko₂ d) k₂o e) none of the above

Explanation:

Step1: Calculate molar - mass of empirical formula

The empirical formula is \(C_{2}H_{4}\). The molar - mass of \(C\) is approximately \(12\ g/mol\) and of \(H\) is approximately \(1\ g/mol\). The molar - mass of \(C_{2}H_{4}\) (\(M_{empirical}\)) is \(2\times12 + 4\times1=24 + 4 = 28\ g/mol\).

Step2: Find the multiple \(n\)

The molecular mass (\(M_{molecular}\)) is given as \(28\ g/mol\). The formula to find the multiple \(n\) between the molecular formula and the empirical formula is \(n=\frac{M_{molecular}}{M_{empirical}}\). Substituting the values, \(n = \frac{28}{28}=1\). Here, the question seems to have a wrong - stated empirical formula for the context of finding \(x\) as the description is not clear. But if we assume the empirical formula \(C_{2}H_{4}\) and molecular formula \((C_{2}H_{4})_x\) and molecular mass \(28\ g/mol\), then \(x = 1\) which is not in the options.

For the second question:

Step1: Assume 100 g of the compound

If we assume 100 g of the compound, then the mass of potassium (\(K\)) is \(69\ g\) and the mass of oxygen (\(O\)) is \(31\ g\).

Step2: Calculate the number of moles

The molar - mass of \(K\) is approximately \(39\ g/mol\) and of \(O\) is approximately \(16\ g/mol\). The number of moles of \(K\), \(n_{K}=\frac{69\ g}{39\ g/mol}\approx1.77\ mol\). The number of moles of \(O\), \(n_{O}=\frac{31\ g}{16\ g/mol}\approx1.94\ mol\).

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles. \(\frac{n_{K}}{n_{K}}\approx1\) and \(\frac{n_{O}}{n_{K}}=\frac{1.94}{1.77}\approx1.1\). Multiply by 2 to get whole - number ratios. We get \(K_2O_2\) which simplifies to \(KO\).

Answer:

1. E. none of the above
2. A. KO