Percent Composition, Empirical Formulas, Molecular Formulas

Percent Composition Percent Composition – the percentage by mass of each element in a compound Percent = _______ Part Whole x 100% Percent composition of a compound or = molecule Mass of element in 1 mol ____________________ Mass of 1 mol x 100%

Percent Composition Example: What is the percent composition of Potassium Permanganate (KMnO 4 )? Molar Mass of KMnO 4 K = 1(39.1) = 39.1 Mn = 1(54.9) = 54.9 O = 4(16.0) = 64.0 MM = 158 g

Percent Composition Example: What is the percent composition of Potassium Permanganate (KMnO 4 )? = 158 g % K Molar Mass of KMnO g K 158 g x 100 =24.7 % % Mn 54.9 g Mn 158 g x 100 = 34.8 % % O 64.0 g O 158 g x 100 =40.5 % K = 1(39.10) = 39.1 Mn = 1(54.94) = 54.9 O = 4(16.00) = 64.0 MM = 158

Percent Composition Determine the percentage composition of sodium carbonate (Na 2 CO 3 )? Molar MassPercent Composition % Na = 46.0 g 106 g x 100% =43.4 % % C = 12.0 g 106 g x 100% =11.3 % % O = 48.0 g 106 g x 100% =45.3 % Na = 2(23.00) = 46.0 C = 1(12.01) = 12.0 O = 3(16.00) = 48.0 MM= 106 g

Percent Composition Determine the percentage composition of ethanol (C 2 H 5 OH)? % C = 52.13%, % H = 13.15%, % O = 34.72% _______________________________________________ Determine the percentage composition of sodium oxalate (Na 2 C 2 O 4 )? % Na = 34.31%, % C = 17.93%, % O = 47.76%

Percent Composition Calculate the mass of bromine in 50.0 g of Potassium bromide. 1. Molar Mass of KBr K = 1(39.10) = Br =1(79.90) =79.90 MM = g ___________ g = x 50.0g = 33.6 g Br 2.

Percent Composition Calculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C 6 H 14 N 2 O Molar Mass of C 6 H 14 N 2 O 2 C = 6(12.01) = H =14(1.01) = MM = g ___________ g = x 85.0 mg = 16.3 mg N 2. N = 2(14.01) = O = 2(16.00) = 32.00

Hydrates Hydrated salt – salt that has water molecules trapped within the crystal lattice Examples: CuSO 4 5H 2 O, CuCl 2 2H 2 O Anhydrous salt – salt without water molecules Examples: CuCl 2 Can calculate the percentage of water in a hydrated salt.

Percent Composition Calculate the percentage of water in sodium carbonate decahydrate, Na 2 CO 3 10H 2 O. 1. Molar Mass of Na 2 CO 3 10H 2 O Na = 2(22.99) = C = 1(12.01) = MM = H = 20(1.01) = 20.2 O = 13(16.00)= H = 20(1.01) = 20.2 Water O = 10(16.00)= MM = g _______ g %x 100%= or H = 2(1.01) = 2.02 O = 1(16.00) = MM H2O = So… 10 H 2 O = 10(18.02) = 180.2

Percent Composition Calculate the percentage of water in Aluminum bromide hexahydrate, AlBr 3 6H 2 O. 1. Molar Mass of AlBr 3 6H 2 O Al = 1(26.98) = Br = 3(79.90) = MM = H = 12(1.01) = O = 6(16.00) = H = 12(1.01) = 12.1 Water O = 6(16.00)= MM = g _______ g %x 100%= or MM = For 6 H2O = 6(18.02) = 108.2

Percent Composition If 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain? MgSO 4. 7 H 2 O 1. Molar Mass Mg = 1 x = g S = 1 x = g O = 4 x = g MM = g H = 2 x 1.01 = 2.02 g O = 1 x = g MM = g MM H 2 O = 7 x g = g Total MM = g g = g 2. % MgSO g g X 100 =48.84 % 3. Grams anhydrous MgSO x 125 =61.1 g

Percent Composition If 145 grams of copper (II) sulfate pentahydrate is completely dehydrated, how many grams of anhydrous copper sulfate will remain? CuSO 4. 5 H 2 O 1. Molar Mass Cu = 1 x = g S = 1 x = g O = 4 x = g MM = g H = 2 x 1.01 = 2.02 g O = 1 x = g MM = g MM H 2 O = 5 x g = 90.1 g Total MM = g g = g 2. % CuSO g g X 100 =63.92 % 3. Grams anhydrous CuSO x 145 =92.7 g

Percent Composition A 5.0 gram sample of a hydrate of BaCl 2 was heated, and only 4.3 grams of the anhydrous salt remained. What percentage of water was in the hydrate? 1. Amount water lost 5.0 g hydrate g anhydrous salt 0.7 g water 2. Percent of water 0.7 g water 5.0 g hydrate x 100 = 14 %

Percent Composition A 7.5 gram sample of a hydrate of CuCl 2 was heated, and only 5.3 grams of the anhydrous salt remained. What percentage of water was in the hydrate? 1. Amount water lost 7.5 g hydrate g anhydrous salt 2.2 g water 2. Percent of water 2.2 g water 7.5 g hydrate x 100 = 29 %

Percent Composition A 5.0 gram sample of Cu(NO 3 ) 2 nH 2 O is heated, and 3.9 g of the anhydrous salt remains. What is the value of n? 1. Amount water lost 5.0 g hydrate g anhydrous salt 1.1 g water 2. Percent of water 1.1 g water 5.0 g hydrate x 100 = 22 % 3. Amount of water 0.22 x =4.0

Percent Composition A 7.5 gram sample of CuSO 4 nH 2 O is heated, and 5.4 g of the anhydrous salt remains. What is the value of n? 1. Amount water lost 7.5 g hydrate g anhydrous salt 2.1 g water 2. Percent of water 2.1 g water 7.5 g hydrate x 100 = 28 % 3. Amount of water 0.28 x =5.0

Formulas Empirical Formula – formula of a compound that expresses lowest whole number ratio of atoms. Molecular Formula – actual formula of a compound showing the number of atoms present Percent composition allow you to calculate the simplest ratio among the atoms found in compound. Examples: C 4 H 10 - molecular C 2 H 5 - empirical C 6 H 12 O 6 - molecular CH 2 O - empirical

Formulas Is H 2 O 2 an empirical or molecular formula? Molecular, it can be reduced to HO HO = empirical formula

Calculating Empirical Formula An oxide of aluminum is formed by the reaction of g of aluminum with g of oxygen. Calculate the empirical formula. 1. Determine the number of grams of each element in the compound g Al and g O 2. Convert masses to moles g Al 1 mol Al g Al = mol Al g O 1 mol O g O = mol O

Calculating Empirical Formula An oxide of aluminum is formed by the reaction of g of aluminum with g of oxygen. Calculate the empirical formula. 3. Find ratio by dividing each element by smallest amount of moles moles Al = mol Al moles O = mol O 4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds) O = x 2 = 3 Al = x 2 = 2 therefore, Al 2 O 3

Calculating Empirical Formula A g sample of cobalt reacts with g chlorine to form a binary compound. Determine the empirical formula for this compound g Co 1 mol Co g Co = mol Co g Cl 1 mol Cl g Cl = mol Cl mol Co mol Cl = 2= 1 CoCl 2

Calculating Empirical Formula When a g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of g. Determine the empirical formula g Fe 1 mol Fe g Fe = mol Fe g O 1 mol O g = mol Fe Fe = g O = g – g = g 1 : 1 FeO

Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains g lead, g of hydrogen, g of arsenic, and g of oxygen. Calculate the empirical formula for lead arsenate g Pb 1 mol Pb g Pb = mol Pb gH 1 mol H g H = mol H g As 1 mol As g As = mol As g Fe 1 mol O g O = mol O

Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains g lead, g of hydrogen, g of arsenic, and g of oxygen. Calculate the empirical formula for lead arsenate mol Pb mol H mol As mol O = mol Pb = 1.00 mol H = mol As = mol O PbHAsO 4

Calculating Empirical Formula The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6. Step 1: In g of Nylon-6 the masses of elements present are g C, g n, 9.80 g H, and g O. Step 2: g C1 mol C g C = mol C g N1 mol N g N = mol N 9.80 g H1 mol H 1.01 g H = 9.72 mol H g O1 mol O g O = mol O

Calculating Empirical Formula The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6. Step 3: mol C = mol C mol N = mol N 9.72 mol H = 11.0 mol H mol O = mol O 6:1:11:1 C 6 NH 11 O

Calculating Molecular Formula A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of g. What is the compound’s molecular formula? Step 1: Molar Mass P = 2 x g = 61.94g O = 5 x 16.00g = g g Step 2: Divide MM by Empirical Formula Mass g g = 2 Step 3: Multiply (P 2 O 5 ) 2 = P 4 O 10

Calculating Molecular Formula A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula? C = g H = 1.01 g g 78 g/mol g/mol = 6 (CH) 6 = C 6 H 6