Give the empirical formula that corresponds to each of the following molecular f

Question

Give the empirical formula that corresponds to each of the following molecular formulas A 0.2999 g sample of a new compound has been analyzed and found to contain the following masses of elements-, carbon. 0.1161 g; hydrogen. 0.02924 g; oxygen. 0.1546 g. Calculate the empirical formula of the compound. If 4.75 g of aluminum metal is heated in a stream of fluorine gas, 14.78 g of aluminum fluoride results. Determine the empirical formula of aluminum fluoride. A compound with the empirical formula CH_2 was found to have a molar mass approximately 72 g. What is the molecular formula of the compound?

Explanation / Answer

1)the empirical formula has the reduced number of atoms of every element.so,

i)KO

ii)C4H3O2

iii)C12H12N2O3

iv)C2H3Cl

2)moles of H=0.02924

moles of oxygen=0.1546/16

=0.0096625

moles of carbon = 0.1161/12

=0.009675

so the empirical formula is CH3O

3)moles of Al=4.75/27

=0.1759

mass of F=14.78-4.75

=10.03

moles of F=10.03/19

=0.528

so the empirical formula is AlF3

4)let the multiplier be x.so,

72=x*(12+2)

or x=5.14

so the formula is C5H10

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