2Simplest formula calculations 3/27/2017Simplest formula calculationsQ- a compound is found to contain the following % by mass: 69.58% Ba, 6.090% C, 24.32% O. What is the simplest (i.e. empirical) formula?Step 1: imagine that you have 100 g of the substance. Thus, % will become mass in grams. E.g % Ba becomes g Ba. (Some questions will give grams right off, instead of %)Step 2: calculate the # of moles (mol = g ¸ g/mol)Step 3: express moles as the simplest ratio by dividing through by the lowest number.Step 4: write the simplest formula from mol ratios.
3Simplest formula: sample problem 3/27/2017Q % Ba, 6.090% C, 24.32% O.What is the empirical (a.k.a. simplest) formula?1: g Ba, g C, g O2: Ba: g ¸ g/mol = mol BaC: g ¸ g/mol = mol CO: g ¸ g/mol = mol O3:mol (reduced)molBaCO1.520/ = 1/ = 1.0011.520/ = 3.0004: the simplest formula is BaCO3
4Mole ratios and simplest formula 3/27/2017Given the following mole ratios for the hypothetical compound AxBy, what would x and y be if the mol ratio of A and B were:A = 1 mol, B = 2.98 molA = mol, B = 1 molA = 2.34 mol, B = 1 molA = 1 mol, B = 1.48 molAB3A4B3A7B3A2B3A compound consists of 29.1 % Na, 40.5 % S, and 30.4 % O. Determine the simplest formula.A compound is composed of 7.20 g carbon, 1.20 g hydrogen, and 9.60 g oxygen. Find the empirical formula for this compoundTry questions on page 189.
5For instructor: prepare molecular models Question 13/27/20171: Assume 100 g: 29.1 g Na, 40.5 g S, 30.4 g O2: Na: g ¸ g/mol = mol NaS: g ¸ g/mol = mol SO: g ¸ g/mol = 1.90 mol O3:mol (reduced)molNaSO1.2661.2631.901.266/ = 1.001.263/ = 11.90/ = 1.504: the simplest formula is Na2S2O3For instructor: prepare molecular models
6Question 23/27/20171: 7.20 g C, 1.20 g H, 9.60 g O2: C: g ¸ g/mol = mol CH: g ¸ 1.01 g/mol = mol HO: 9.6 g ¸ g/mol = 0.60 mol O3:mol (reduced)molCHO0.59951.1880.600.5995/ = 11.188/ = 1.980.60/ = 1.04: the simplest formula is CH2O
7Question 3 1: Assume 100 g: 28.9 g K, 23.7 g S, 47.7 g O 3/27/20171: Assume 100 g: 28.9 g K, 23.7 g S, 47.7 g O2: C: 7.20 g ¸ g/mol = mol CH: 1.20 g ¸ 1.01 g/mol = mol HO: 9.6 g ¸ g/mol = 0.60 mol O3:mol (reduced)molCHO0.59951.1880.600.5995/ = 11.188/ = 1.980.60/ = 1.04: the simplest formula is CH2O
8Molecular formula calculations 3/27/2017There is one additional step to solving for a molecular formula. First you need the molar mass of the compound. E.g. in Q2, the molecular formula can be determined if we know that the molar mass of the compound is 150 g/mol.First, determine molar mass of the simplest formula. For CH2O it is 30 g/mol ( ).Divide the molar mass of the compound by this to get a factor: 150 g/mol ¸ 30 g/mol = 5Multiply each subscript in the formula by this factor: C5H10O5 is the molecular formula. (models)Q- For OF, give the molecular formula if the compound is 70 g/molO2F ¸ 35 = 2
9Combustion analysis gives the following: 3/27/2017Combustion analysis gives the following:26.7% C, 2.2% hydrogen, 71.1% oxygen.If the molecular mass of the compound is g/mol, determine its molecular formula.What information must be known to determine a) the empirical formula of a substance? b) the molecular formula of a substance?A compound’s empirical formula is CH, and it weighs 104 g/mol. Give the molecular formula.A substance is decomposed and found to consist of 53.2% C, 11.2% H, and 35.6% O by mass. Calculate the molecular formula of the unknown if its molar mass is 90 g/mol.
10Question 7 1: Assume 100 g total. Thus: 3/27/20171: Assume 100 g total. Thus:26.7 g C, 2.2 g H, and 71.1 g O2: C: g ¸ g/mol = mol CH: 2.2 g ¸ 1.01 g/mol = 2.18 mol HO: g ¸ g/mol = mol O3:2.2232.184.4442.223/2.18 = 1.022.18/ = 14.444/ = 2.044: the simplest formula is CHO25: factor = 90/45=2. Molecular formula: C2H2O4
113/27/2017Question 8, 9For the empirical formula we need to know the moles of each element in the compound (which can be derived from grams or %).For the molecular formula we need the above information & the molar mass of the compoundMolar mass of CH = 13 g/molFactor = 104 g/mol ¸ 13 g/mol = 8Molecular formula is C8H8
12Question 10 1: Assume 100 g total. Thus: 3/27/20171: Assume 100 g total. Thus:53.2 g C, 11.2 g H, and 35.6 g O2: C: g ¸ g/mol = mol CH: g ¸ 1.01 g/mol = mol HO: g ¸ g/mol = mol O3:4.43011.092.2254.43/ = 1.9911.09/ = 4.982.225/2.225 = 14: the simplest formula is C2H5O5: factor = 90/45=2. Molecular formula: C4H10O2
133/27/2017AssignmentCalculate the percentage composition of each substance: a) SiH4, b) FeSO4Calculate the simplest formulas for the compounds whose compositions are listed:a) carbon, 15.8%; sulfur, 84.2%b) silver,70.1%; nitrogen,9.1%; oxygen,20.8%c) K, 26.6%; Cr, 35.4%, O, 38.0%The simplest formula for glucose is CH2O and its molar mass is 180 g/mol. What is its molecular formula?
14percentage composition molar mass(g/mol) 64.9% C, 13.5% H, 21.6% O 74 Determine the molecular formula for each compound below from the information listed.substance simplest formula molar mass(g/mol)a) octane C4H9 114b) ethanol C2H6O 46c) naphthalene C5H4 128d) melamine CH2N2 126The percentage composition and approximate molar masses of some compounds are listed below. Calculate the molecular formula of eachpercentage composition molar mass(g/mol)64.9% C, 13.5% H, 21.6% O 7439.9% C, 6.7% H, 53.4 % O 6040.3% B, 52.2% N, 7.5% H 803/27/2017
15the simplest formula is CS2 3/27/20171 a) Si= 87.43% (28.09/32.13 x 100), H= 12.57%b) Fe= 36.77% (55.85/ x 100), S= 21.10% (32.06/ x 100), O= 42.13%2 a) Assume 100 g. Thus: 15.8 g C, 84.2 g S.C: 15.8 g ¸ g/mol = mol CS: 84.2 g ¸ g/mol = mol S2.626/1.315= 2.001.315/1.315= 1Mol reduced2.6261.315MolSCthe simplest formula is CS2
162 b) Ag: 70.1 g ¸ g/mol = mol AgN: 9.1 g ¸ g/mol = mol NO: 20.8 g ¸ g/mol = 1.30 mol O3/27/2017.6499/.6495= 1.00.6499Ag1.30/.6495= 2.00.6495/.6495= 1Mol reduced1.300.6495MolONAgNO22 c) K: 26.6 g ¸ g/mol = mol KCr: g ¸ g/mol = mol CrO: 38.0 g ¸ g/mol = mol O.6803/.6803= 10.6803K2.375/.6495= 3.49.6808/.6803= 1.00Mol reduced2.3750.6808MolOCrK2Cr2O7