Calculate percent composition of KNO3

• KNO3

• Molar mass = 101.1 g/molPotassium: (39.10 / 101.1) x 100 = 38.67%Nitrogen: (14.01 / 101.1) x 100 = 13.86%Oxygen: (48.00 / 101.1) x 100 = 47.48%

• How much K would be expected from 2000. kg of KNO3?• 38.67 kg K x 2000. kg = 773.4 kg 100 kg KNO3

A Poem

Percent to massMass to mole

Divide by smallMultiply 'til whole

That gets you empirical!Multiply subscripts

‘Til the mass matches given molar massThat gets you molecular!

• Percent to massMass to moleDivide by smallMultiply 'til whole

Here's an example of how it works. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?

• (1) Percent to mass:Assume 100 g of the substance, then 72.2 g magnesium and 27.8 g nitrogen.

• (2) Mass to moles:for Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) = 2.97 mol Mgfor N: 27.8 g N x (1 mol N/14.0 g N) = 1.99 mol N(3) Divide by small:for Mg: 2.97 mol / l.99 mol = 1.49for N: 1.99 mol / l.99 mol = 1.00(4) Multiply 'til whole:for Mg: 2 x 1.49 = 2.98 (i.e., 3)for N: 2 x 1.00 = 2.00and the formula of the compound is Mg3N2.

More details

(1) Percent to mass:• Assume 100 g of the substance, then 72.2 g

magnesium and 27.8 g nitrogen. Percent to mass: the assumption of 100 grams is purely for convenience sake. This means that the percentages transfer directly into grams. If you assumed that 36.7 grams were present, you would have to multiply 36.7 by each percentage. Assuming 100 grams makes it much easier.

(2) Mass to moles:• for Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) =

2.97 mol Mgfor N: 27.8 g N x (1 mol N/ 14.0 g N) = 1.99 mol NMass to moles: this is a technique you should ALREADY know. If you don’t, take the time to learn it.

(3) Divide by small:for Mg: 2.97 mol / 1.99 mol = 1.49for N: 1.99 mol / 1.99 mol = 1.00Divide by small: make sure you divide ALL answers from #2 by the smallest value. This may seem obvious, but lots of students forget this step.

Mg: 2.97 mol / 1.99 mol = 1.49N: 1.99 mol / 1.99 mol = 1.00

(4)Multiply 'til whole:Mg 1.49 is closer to 1.5, decimal equivalent to ½, than either 1 or 2. Don’t round it to a whole number! Multiply both values by denominator 2 to get whole numbers.

for Mg: 2 x 1.49 = 2.98 (i.e., 3) for N: 2 x 1.00 = 2.00

Multiply 'til whole: multiply ALL values from #3 by the same factor. This factor is selected so as to produce ALL whole numbers as answers. Look for decimal equivalents of fractions!

• and the formula of the compound is Mg3N2.

Example: A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. The molecular weight of this compound is known to be approximately 140 g/mol. What is the empirical formula? What is the molecular formula?

• (1) Percent to mass. Assume 100 grams of the substance is present; % gcarbon: 68.54 gramshydrogen: 8.63 gramsoxygen: 22.83 grams

• (2) Mass to moles. Divide each mass by the proper atomic weight.carbon: 68.54 / 12.011 = 5.71 molhydrogen: 8.63 / 1.008 = 8.56 moloxygen: 22.83 / 16.00 = 1.43 mol

• (3) Divide by small. Divide each value by the smallest to try to get whole number ratiocarbon: 5.71 ÷ 1.43 = 3.99hydrogen: 8.56 ÷ 1.43 = 5.99oxygen: 1.43 ÷ 1.43 = 1.00

• (4) Multiply 'til whole. Not needed since all values came out very close to whole numbers

• The empirical formula of the compound is C4H6O.• But we’re not done…Molar mass is given in the problem. Does it match the

empirical formula mass?

Multiply subscripts to match given mass

• The empirical formula of the compound is C4H6O.• (1) Calculate the "empirical formula mass."

C4H6O gives an "EFM" of 70.092.• (2) Divide the molecular mass by the "EFM."

140 ÷ 70 = 2• (3) Multiply the subscripts of the empirical formula

by the factor just computed.C4H6O times 2 gives a formula of C8H12O2.This is the molecular formula.