II. Formula Calculations Ch 10 The Mole - Formula Calculations Ch. 10 – The Mole II. Formula Calculations Chilton

What is Percent Composition?

Ch 10 The Mole - Formula Calculations A. Percent Composition the percentage by mass of each element in a compound Use this formula when you are given the masses of the individual elements (Ex. 1) Use this formula when you are only given the chemical formula (Ex. 2) Chilton

A. Percent Composition Ex. 1) Find the percent composition of a sample that is 28 g Fe and 8.0 g O. Known: Mass of Fe = 28 g Mass of O = 8.0 g Total Mass = 28 + 8.0 g = 36 g Unknown: % Fe = ? % O = ? 28 g 36 g Check: 78% + 22% = 100% %Fe =  100 = 78% Fe 8.0 g 36 g %O =  100 = 22% O

A. Percent Composition  100 = 79.86% Cu %Cu =  100 = 20.14% S %S = Ex. 2) Find the % composition of Cu2S. Use this formula when finding % composition from a chemical formula: Known: Mass of Cu in 1 mol Cu2S = 2(63.55g) = 127.10 g Cu Mass of S in 1 mol Cu2S = 32.06 g S Molar Mass = 127.10 g + 32.06 g = 159.16 g/mol Unknown: % Cu = ? % S = ? 127.10 g Cu 159.16 g Cu2S  100 = 79.86% Cu %Cu = 32.06 g S 159.16 g Cu2S  100 = 20.14% S %S =

A. Percent Composition How many grams of copper are in a 38.0-gram sample of Cu2S? Use answer from last question Cu2S is 79.86% Cu (38.0g)(.7986) = 30.3 g Cu

A. Percent Composition Find the percent composition of Cu2SO4. Known: Mass of Cu in 1 mol Cu2SO4 = 2(63.55 g) = 127.10 g Cu Mass of S in 1 mol Cu2SO4 = 32.06 g S Mass of 0 in 1 mol Cu2SO4 = 4(16.00 g) = 64.00 g O Molar Mass = 127.10 g + 32.06 g + 64.00 g = 223.16 g/mol Unknown: % Cu = ? % S = ? % O = ?

A. Percent Composition Find the percent composition of Cu2SO4. Check: 127.10 g 223.16 g %Cu =  100 = 56.95% Cu Check: 56.95% + 14.37% + 28.68% = 100% 32.06 g 223.16 g  100 = 14.37% S %S = 64.00 g 223.16 g %O =  100 = 28.68% O

em·pir·i·cal adjective Definition of 1: originating in or based on observation or experience < data> 2: relying on experience or observation alone often without due regard for system and theory <an empirical basis for the theory>

C2H6 CH3 B. Empirical Formula Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts CH3

B. Steps to Find Empirical Formula 1. Find mass (or %) of each element. 2. Calculate moles of each element. 3. Divide moles (from #2) by the smallest # of moles to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

B. Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g N 1 mol N 14.01 g N = 1.85 mol N = 1 N 1.85 mol N1.85O4.63 74.1 g O 1 mol O 16.00 g O = 4.63 mol O = 2.5 O

N2O5 N1O2.5 B. Empirical Formula Need to make the subscripts whole numbers  multiply by 2 N2O5

EF = OH B. Empirical Formula Find the empirical formula for a sample of 94.1% O and 5.90% H. 94.1 g O 1 mol O 16.00 g O = 5.88 mol O = 1 O 5.84 mol 5.9 g H 1 mol H 1.01 g H = 5.84 mol H = 1 H EF = OH

EF = Mn2O7 B. Empirical Formula = 0.9013 mol Mn = 1 Mn = 3.154 mol O #4. An oxide of manganese contains 49.52% manganese and 50.47% oxygen. What is its empirical formula? 49.52 g Mn 1 mol Mn 54.94 g Mn = 0.9013 mol Mn = 1 Mn 0.9013 mol x 2 50.47 g O 1 mol O 16.00 g O = 3.154 mol O = 3.50 O x 2 EF = Mn2O7

CH3 C2H6 C. Molecular Formula “True Formula” – the actual number of atoms in a compound, either the same as or a whole-number multiple of the empirical formula CH3 empirical formula ? C2H6 molecular formula

C. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula MM. 3. Divide the true molar mass by the empirical formula MM. 4. Multiply each subscript in your EF by the answer from step 3.

(EF)n = (CH2)2  C2H4 C. Molecular Formula The empirical formula for ethylene is CH2. Find the molecular formula if the molar mass is 28.1 g/mol! empirical formula mass = 14.03 g/mol 28.1 g/mol 14.03 g/mol n = = 2.00 (EF)n = (CH2)2  C2H4

D. Put it all together!! 1,6-diaminohexane is 62.1% C, 13.8% H, and 24.1% N. What is the empirical formula? If the molar mass is 116.21 g/mol, what is the molecular formula? 62.1 g C 1 mol C 12.01 g C = 5.17 mol C = 3 C 1.72 mol 13.8 g H 1 mol H 1.01 g H = 13.7 mol H = 8 H 24.1 g N 1 mol N 14.01 g N = 1.72 mol N = 1 N EF = C3H8N

C3H8N C6H16N2 (C3H8N)2  D. Put it all together!! empirical formula mass = 58.12 g/mol 116.21 g/mol 58.12 g/mol n = = 2.00 C6H16N2 (C3H8N)2 