Question

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STOIO388: The Empirical Formula of an Oxide Ad) weighound is 303 respectively. Det 13728, On analysis it was found to and is 303.8 g mol and the molar 1 and 107.9 g mol, respectively. Determine 2. A sample of a compound containing C, O, and silver (Ag) weighed 1.372 g. On analysis it was contain 0.288 g O and 0.974 g Ag. The molar mass of the compound is 303.8 g mol-1 and masses of Co, and Ag are 12.01 g mol-16.00 g mol, and 107.9 g mol-, respectively the empirical and molecular formulas of the compound. 3. Briefly explain how your Mg to Oratio would have been affected if you had failed to convert all of the Mg N2 present to magnesium oxide in Steps 18-22 of the procedure.

Answer 1

2. Total weight of the sample = 1.372 g

weight of Oxygen = 0.288 g

weight of silver = 0.974 g

weight of carbon = 1.372 - ( 0.288 + 0.974 ) = 0.11 g

ELEMENT	GIVEN WEIGHT	ATOMIC WEIGHT	RATIO =GIVEN WEIGHT/ATOMIC WEIGHT	SIMPLEST RATIO
Carbon	0.11	12.01	0.11/12 = 0.0091	1
Oxygen	0.288	16	0.288/16 = 0.018	2
Silver	0.974	107.90	0.974/107.90 = 0.009	1

Simplest ration is found by dividing each ratio with 0.009

empirical formula = AgCO₂

empirical mass = 107.9 +12.01 +(2 x 16) = 151.91

but given molecular mass = 303.8

n = (molecular mass/ empirical mass)

n = 303.8/ 151.91 = 2

$\therefore$ Molecular formula = n x empirical formula

molecular formula = 2 x AgCO₂

molecular formula = Ag₂C₂O₄