# Empirical and Molecular Formulas

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Empirical and Molecular Formulas

Definition: the lowest whole number ratio of elements in a compound.
Steps to determine empirical formulas: 1) find moles of each element [find the moles of H20 if compound is hydrated]. 2) divide by smallest number of moles to get a whole number. 3) multiply by a common whole number, if necessary.

Calculate EF of a compound with 25.9% N and 74.1% O.
%=assume mass mass mols 25.9g N x 1 mol N = mols N 14.007g N 74.1g O x 1 mol O = 4.63 mols O 16 g O 2) N: = 1 1.85 O: = 2.5 3) Multiply by common whole number N: 1 x 2 =2 O: 2.5 x 2 = = N2O5

Just remember this!! Percents to grams Grams to moles Divide by smallest Multiply ‘til whole

Given: 36.5g Na, 25.4g S, and 38.1g O 25.4g S x 1 mol S = .79 mols S
36.5g Na x 1 mol Na = 1.59 mols Na 22.989g Na 25.4g S x 1 mol S = .79 mols S 32g S 38.1g O x 1 mol O = 2.4 mols O 16g O Na: = S: .79 = O: 2.4 = 3 = Na2SO3

An oxide of aluminum is formed by the reaction of 4
An oxide of aluminum is formed by the reaction of 4.151g of aluminum with 3.692g of oxygen. Find EF. 4.151g Al x 1 mol Al = mol Al atoms 26.98 g Al 3.692g O x 1 mol O = mol O atoms 16 g O mol Al = mol Al atoms 0.1539 mol O = mol O atoms 1.500 O x 2 = = 3 O atoms 1.000 Al x 2 = 2.000= 2 Al atoms = Al2O3

Multiple = molecular mass x empirical formula (whole number)
Molecular formulas show the number of atoms in a compound. In order to determine the molecular, you must have the empirical formula first. The molecular mass of molecule will always be given. Multiple = molecular mass x empirical formula (whole number) The molecular formula is always an integer multiple of the empirical formula.

Determine the molecular formula of a compound whose EF is CH2O and molecular mass is 120 g/mol
-Distribute that 4 throughout the empirical formula = C4H8O4

A compound is 64. 9% carbon, 13. 5% hydrogen, and 21. 6% oxygen
A compound is 64.9% carbon, 13.5% hydrogen, and 21.6% oxygen. Its molc mass is 74 g/mol. What is its MF? 64.9g C x 1 mol C = 5.40 mols C = 4 C 12.01 g C 13.5g H x 1 mol H = mols H = 10 H 1.01 g H 21.6g O x 1 mol O = 1.35 mols O = 1 O 16.0 g O 74 g/mol = 1 74 g/mol = C4H10O

Ex) a compound is 54. 5% carbon, 9. 1% hydrogen, and 36. 4% oxygen
Ex) a compound is 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen. It’s molc mass is 88 g/mol. What is its molecular formula? 54.5 C x 1 mol C = 4.54 mols C 12.01 g C 9.1 g H x 1 mol H = 9 mols H 1.01 g H 36.4 g O x 1 mol O = 2.28 mols O 16 g O 4.54 mols C = 2 C 2.28 mols 9 mols H = 4H 2.28 mols O = 1O = C4H8O2

Ex) A compound has an empirical formula of ClCH2 and a molecular weight of g/mol. What is it’s molecular formula? Mass Cl + mass C + 2(mass H) Mass of empirical unit= (2.008) = g/mol 98.96 g/mol = 2.000 49.47 g/mol = Cl2C2H4