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What is the molecular formula for the compound containing 30.4 percent nitrogen and 69.6 percent oxygen by mass and having a molar mass of 92.0 g?
Wiki User
∙ 11y ago
Assume those percentages as mass. Find moles.
30 grams Nitrogen (1 mole N/14.01 grams) = 2.141 moles ( call it 2 )
70 grams oxygen (1 mole O/16.0 grams) = 4.4 moles ( call it 4 )
smallest mole number is 1 in formula and divide larger number by it
2/4 = 2
so..............
NO2
nitrogen dioxide is the empirical formula
Wiki User
∙ 13y ago
Wiki User
∙ 10y agoELEMENT % ATOMIC RELATIVE NO. SIMPLEST SIMPLEST
MASS OF ATOMS RATIO whole no.ratio
Iron 69.9% 26 69.9/26 2.69/2.69 1*5=5
=2.69 =1
oxygen 30.1% 8 30.1/8 3.76/2.69 1.4*5=7
=3.76 =1.4
So, The empirical formula is Fe5O7
Wiki User
∙ 11y agoDivide each percentage by the Atomic Mass of that element.
Nitrogen: 30.4 / 14 = 2.1714
Oxygen: 69.6 / 16 = 4.35
compare these and divide by the smallest, round if necessary.
Nitrogen: 2.1714 / 2.1714 = 1
Oxygen: 4.35 / 2.1714 round to 2
This gives an empirical formula of NO2
The mass of this empirical formula = 14 + 2x16 = 46
Divide the molar mass by this empirical mass. 92.0 / 46 = 2
multiply the subscripts in the empirical formula by this number to get the molecular formula
(NO2)2 = N2O4 THIS IS THE MOLECULAR FORMULA.
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