Chemistry 1 (CHM 2045 )-19 PDF

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These notes include lecture notes, lab assignment information and examples of how to do problems plus it has some quiz questions....

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Lecture 8 interconverting mass and number of particles

Example • How many copper atoms are in a pure copper penny with a mass of 3.10 g? • Strategy: convert mass to moles using the molar ma of copper as a conversion factor, then moles to atoms using Avogadro’s number. AtomsCu=3.10gCu× 1molCu ×6.022×1023 atoms =2.94

× 1022 atoms Cu

• How much does 1.67 ×1023 water molecules weigh? • Strategy: convert molecules to moles using Avogadro’s number, then moles to mass using the molar mass of water as a conversion factor. • First, calculate the molar mass of water: Molar mass of H2O = (2 × 1.01 g/mol) + (1 × 16.00 g/mol) Molar mass of H2O = 18.02 g/mol Mass (g) = 1.67 × 1023 molecules × 1 mol × 18.02 g H2O

= 5.00 g H2O (about 1 tsp)

Quiz questions

• How many magnesium atoms are there in a strip of pure magnesium that weighs 1.69 g? (b) 2.47 × 1025 atoms (c) 1.16 × 10-25 atoms (d) 8.66 × 1024 atoms (e) TAP OUT • What is the mass of 2.25 × 1022 tungsten atoms? (a) 6.76 g (b) 4.78 g (c) 1.79 g (

(e) TAP OUT

• How many moles of uranium are in one mole of autunite, Ca(UO2)2(PO4)2? (a) 1⁄2

(e) TAP OUT • What is the mass of oxygen in 3.2 g of CaCO3? (a) 3.2 g ( OUT

(

g(

9.6 g (e) TAP

• What is the empirical formula for the unknown compound with percent composition 18.1 % C, 2.27% H and 79.8 % Cl? (a) CHCl (b) C2H2Cl (c) CH2Cl

3

(e) TAP OUT

Quiz questions • What is the molecular formula for a compound with percent composition 44.44 % C, 3.73 % H and 51.83 % N and molecular mass = 135.13 g/mol? (a) CNH (b) C2H2N (c) CHN2

(e) TAP OUT

• Combustion analysis of an unknown hydrocarbon shows the products are 1.320 g CO2 and 0.540 g H2O. Given that mass spectrometry finds the hydrocarbon has a molecular mass of 84.0 g/mol, what is its molecular formula? (a) CH (b) C6H6 (d) CH2 (e) TAP OUT

How much a given element is in a compound or mixture • First, we need to relate moles of given element to moles of compound • A chemical formula tells us the number of atoms per mole of a molecular or ionic compound. • Therefore, for each mole of a compound, the subscript denoting the number of a individual atom gives us the mole ratio for moles of element to moles of compound. e.g. 1 mol H2O molecules = 6.02 × 1023 H2O molecules 1 mol H2O molecules contains 2 mol H atoms = 2(6.02 × 1023) H atoms and 1 mol O atoms = 6.02 × 1023 O atoms Example

• How many moles of uranium are in one mole of uranophane, CaU2Si2O11? • Strategy: examine the chemical formula of uranophane to work out the conversion factor between moles of U and moles of CaU2Si2O11. • For every 1 mole of CaU2Si2O11 there are 2 moles of U atoms. • Because 1 mol = 6.022 × 1023 particles, for every 6.022 × 1023 CaU2Si2O11 formula units there has to be 2 × (6.022 × 1023) = 1.204 × 1024 U atoms.

Calculating mass of an element in a compound or mixture 1. Use the mass and molar mass of the compound to calculate the number of moles of compound present. 2. Use the mole ratio of moles of element to moles of compound as a conversion factor. 3. Use the molar mass of the element to calculate the mass of the element within the compound. Example

• What is the mass of hydrogen in 5.0 g of water? • Strategy: examine the chemical formula of water to work out the conversion factor between moles of H and moles of H2O. • For every 1 mole of H2O there are 2 moles of H atoms so the conversion factor is 2 mol H Molar mass of H2O = (2 × 1.01 g/mol) + (1 × 16.00 g/mol) = 18.02 g/mol MassH(g)=5.0gH2O× 1molH2O × 2molH ×1.01gH

= 0.55 g

Calculating empirical formula from percent composition 1. Assume you have 100 g of the substance so percent composition values are equivalent to mass in g. 2. Convert mass to moles for each element using molar mass. 3. Divide moles of each element by the smallest number of moles to convert to a mole ratio. 4. If necessary, convert the mole ratio to the smallest ratio of whole numbers.

Example 3. Divide moles of each element by the smallest number of moles to convert to a mole ratio. Mol ratio Cr = 1.317307692 mol Cr = 1 mol Cr Mol ratio O = 1.96875 mol O = 1.5 mol O 4. If necessary, convert the mole ratio to the smallest ratio of whole numbers. 1 mol Cr:1.5 mol O therefore 2 mol Cr:3 mol O. Empirical formula = Cr2O3

Comparing empirical and molecular formula • Empirical formula is the same as a formula unit for an ionic compound. • Recall that molecular formula shows the number and type of atoms present in one molecule of a molecular compound. • Empirical formula is not always the same as molecular formula for a molecular compound! • However, molecular formula should always be related to empirical formula by a multiplier n: n = molecular mass of molecular compound Example • A molecular compound has the empirical formula PNCl2 and a molecular mass = 579.43 g/ mol, what is its molecular formula? • Strategy: •

Calculate the molar mass of empirical formula PNCl2

•

Divide the molecular mass of the compound by the molar mass of the empirical formula find n.

•

Multiply the empirical formula by n to find the molecular formula.

Example 2

• MolarmassPNCl2 =(1×30.97)+(1×14.01)=(2× 35.45) = 115.88 g/mol • Find n: n = molecular mass of molecular compound n = 579.43 g/mol = 5 • Multiply empirical formula by n to find molecular formula: (PNCl2)n = (PNCl2)5 = P5N5Cl10

• What is the molecular formula for a compound with percent composition 62.58 % C, 9.63 % H and 27.75 % O and molecular mass = 230.30 g/mol? 1. Assume you have 100 g of the substance so percent composition values are equivalent to mass in g. Mass C = 62.58 g Mass H = 9.63 g Mass O = 27.75 g 2. Convert mass to moles for each element using molar mass. MolC=62.58gC× 1molC =5.210657785mol MolH=9.63gH× 1molH =9.534653465mol MolO=27.75gO× 1molO =1.734375mol

Example 3

3. Divide moles of each element by the smallest number of moles to convert to a mole ratio. Mol ratio C = 5.210657785 mol C = 3 mol C Mol ratio H = 9.534653465 mol H = 5.5 mol H Mol ratio O = 1.734375 mol O = 1 mol O 4. If necessary, convert the mole ratio to the smallest ratio of whole numbers. 3 mol C:5.5 mol H:1 mol O 6 mol C:11 mol H:2 mol O Empirical formula = C6H11O2

• MolarmassC6H11O2 = (6 × 12.01) + (11 × 1.01) + (2 × 16.00) = 115.17 g/mol • Find n: n = molecular mass of molecular compound n = 230.30 g/mol = 2

• Multiply empirical formula by n to find molecular formula: (C6H11O2)n = (C6H11O2)2 = C12H22O4

Combustion Analysis

• Combustion analysis burning a substance completely in oxygen to produce compounds whose masses are measured to determine the composition of the original material. • Carried out in excess oxygen (more than the stoichiometric amount) to ensure complete combustion. • General reaction for burning a hydrocarbon: CaHb + excess O2(g) → aCO2(g) + b2H2O(g) • We can calculate the empirical formula of the hydrocarbon by finding the mole ratio of C and H. 1. Use the mass of CO2 released to calculate the number of moles of CO2 and thus the number of moles of C in the hydrocarbon. 2. Use the mass of H2O released to calculate the number of moles of H2O and thus the number of moles of H in the hydrocarbon. NOTE: 1 mol H2O contains 2 mol H atoms! (Optional step if third element present in compound: Calculate the mass of C and H in the products and subtract this from the mass of the original sample to find the mass of the third element.) 3. Divide both mole amounts of C and H by the smaller one to find the mole ratio of C to H and thus the empirical formula of the hydrocarbon. 4. If we know the molecular mass (from mass spectrometry) we can find the molecular formula.

Combustion analysis example...