Download Chemistry: Quantitative Analysis and Percentage Composition Calculations and more Summaries Chemistry in PDF only on Docsity! No Brain Too Small @ CHEMISTRY 3€ Achievement Standard Chemistry 91161: Carry out quantitative analysis In addition to the titration (see other resources on No Brain Too Small), this AS also has a written component with calculations. Important formula: n = m/M n=m/M (which can be rearranged to m=nxM m where nis the amount of substance (mol) n|M m is the mass of substance (g) and M is the molar mass of the substance (g mol?) The portion of the periodic table shown below shows the atomic number and molar mass of some elements. E.g. Na has a molar mass of 23.0 g mol”. This means 1 mol of Na has a mass of 23.0 g. This can be written M(Na) = 23.0g mol in questions. Atomic number [T H 1 2 1.0 _|Molarmass/g mot! 3 4 ui Be 69 | 90 I |a2 Na | Mg 230 | 243] 3 4 5 6 7 8 S10 19 20 21 22 23 24 25 26 27 28 K Ca Se Ti v Cr | Mn Fe Co NL 39.1 | 401 | 450 | 479 | so9 | 20 | s49 | 359 | 58.9 | 58.7 alae Ta sdTansdan—sday—sdazsdaa—sdas saw What is the mass of 1 mol of Mg? Answer: 24.3 g What is the mass of 2 mol of Mg? Answer: m=nxM m=2x 24.3 =48.6g What is the mass of 0.75 mol of Mg? Answer: m=nxM m = 0.75 x 24.3 = 18.225 g How many mol of Mg are there is 56.0 g of Mg? Answer: n =m/M n = 56.0/24.3 = 2.30 mol % Composition calculations Citric acid, formula CsHgO7, can be used to soften water which makes it useful in soaps and laundry detergents. It is also the acid found in citrus fruits. Calculate the percentage composition of citric acid. M(C) = 12.0 g mol? M(H) = 1.00 g mol? M(O) = 16.0 g mol + First calculate (M) CsHg07 © (M) CeHgO7 = (12.0 x 6) + (1.00 x 8) + (16.0 x 7) = 192 g mol? © =%C= (12.0 x 6) x 100 = 37.5% 192 These will add ° 7H = (1.00 x8) «100 = 4.17% up to 200% of © =%0 = (16.0 x 7) x 100 = 58.3% course © 192 No Brain Too Small @ CHEMISTRY 3€ Rhubarb contains acids, one of which is malic acid, formula CaHeOs. Calculate the percentage composition of malic acid. First calculate (M) CgHeOs © (M)CaHeOs = (12.0 x 4) + (1.00 x 6) + (16.0 x 5) = 134 g mol * © =%C= (12.0 x 4) x 100 = 35.8% 134 © %H = (1.00 x 6) x 100 = 4.48% 134 — © =%0 = (16.0 x 5) x 100 = 59.7% 134 Empi | formula and the molecular formulae Empirical formulas give the lowest whole number ratio of the atoms in a compound. The molecular formula gives the exact composition of one molecule. Examples: molecular formula empirical formula CH, CH, C6H120¢6 CH20 H20 H20 CaHio Hs Example 1: Rhubarb plants contain oxalic acid. Oxalic acid is 26.7% carbon, 2.22% hydrogen and 71.1% oxygen, and has a molar mass of 90.0 g mol. Use the following molar masses to calculate both the empirical formula and the molecular formula of oxalic acid: M(C) = 12.0 g mol"? M(H) = 1.00 g mol? M(O) = 16.0g mol? e Assume that you have 100 g of the compound. Then 100 g of oxalic acid would contain 26.7 g of carbon, 2.22 g of hydrogen and 71.1 g of oxygen Carbon Hydrogen Oxygen Mass (g) 26.7 2.22 71.1 Moles (mol) 26.7 / 12.0 2.22/1.00 71.1/16.0 n=m/M = 2.225 = 2.22 = 4.44375 divide each by 2.225/2.22 2.22/2.22 4.44375/2.22 the smallest # of = 1.00 = 1.00 = 2.00 moles Simplest whole 1 1 2 number ratio © Ratio is 1:1:2 so empirical formula is CHO Oxalic acid has a molar mass of 90.0 g mol 7. e Empirical formula CHO2 has a molar mass of 12.0 + 1.00 + (16.0 x 2) = 45.0g molt. e #of CHO: units is 90.0/45.0 = 2 e Molecular formula is therefore 2 x CHO2, that is Cp2H2O4
