Question

# A sample of a compound containing C, O, and silver (Ag) weighted1.372 g. On analysis it...

A sample of a compound containing C, O, and silver (Ag) weighted1.372 g. On analysis it was found to contain 0.288 g O and 0.974 gAg. The molar mass of the compound is 303.8 g mol-1andthe molar masses of C, O, and Ag are 12.01 g mol-1,16.00 g mol-1, and 107.9 g mol-1,respectively. Determine the empirical and molecular formulas of the compound .

First find the mass of Carbon

mass of C = 1.372 - 0.288 - 0.974 = 0.11 g

Then calculate the moles of each substance dividing each mass with the Atomic Weight of the element:

Ag: 0.974/108 = 9*10^(-3) mol
C: 0.11/12 = 9*10^(-3) mol
O: 0.288/16 = 18*10^(-3) mol

Now divide by the smallest value:

Ag: 9*10^(-3)/9*10^(-3) = 1 mol
C: 9*10^(-3)/9*10^(-3) = 1 mol
O: 18*10^(-3)/9*10^(-3) = 2 mol

So the empirical formula is: (AgCO2)v

Evaluate the molar mass:

Mr = (108 + 12 + 2*16)v = 152v

But Mr = 308, so 152v = 308, v = 2 approx. So the molecular formula of the substance is:

Ag2C2O4 (probably is the silver oxalic salt: AgOOC-COOAg)

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