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An unknown compound has a percent composition of 52.10% potassium, 15.8% carbon, and 32.1% oxygen. The molar mass of the compound is 150.22 g/mol. What is the empirical formula and the molecular formula of this compound?

Respuesta :

A) KCO and K2C2O3
B) K2C2O3 and K6C6O9
C) K2C2O3 and K2C2O3
D) K2C2O3 and K4C4O6
E) K2C2O3 and KCO1.5

take the full amount, then multiply by percentage as a decimal

[potassium]150.22*0.521= 78....

[Carbon]150.22*0.158=23...

[Oxygen]150.22*0.321=48...

Then divide these by their Molar Mass.

78../39.1=2moles

23.../12.0=1.97.. moles

48.../16= 3 moles

so molecular formula is K2C2O3, and therefore empirical is the same, cant be simplified anymore

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