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2 Chapter 10 Chemical Quantities or

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3 How you measure how much? How you measure how much? n You can measure mass, n or volume, n or you can count pieces. n We measure mass in grams. n We measure volume in liters. n We count pieces in MOLES.

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4 Moles n 1 mole is 6.02 x particles. n Particles can be atoms, molecules, formula units. n 6.02 x is called Avogadro's number.

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5 Representative particles n The smallest pieces of a substance. n For an element it is an atom. –Unless it is diatomic n For a molecular compound it is a molecule. n For an ionic compound it is a formula unit.

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6 Conversion factors n Used to change units. n Three questions –What unit do you want to get rid of? –Where does it go to cancel out? –What can you change it into?

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7 Calculation question n How many molecules of CO 2 are the in 4.56 moles of CO 2 ? 4.56 moles CO 2 x 6.02 x Molecules = 2.75 x Molecules CO 2 1 Mole CO 2 1 Mole CO 2

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8 Calculation question n How many moles of water is 5.87 x molecules? 5.87 x molecules H 2 O x 1 Mole H 2 O____________ =.098 Moles H 2 O 6.02 x Molecules H 2 O 6.02 x Molecules H 2 O

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9 Calculation question n How many atoms of carbon are there in 1.23 moles of C 6 H 12 O 6 ? –1.23 moles C 6 H 12 O 6 x 6 moles C Atoms_____ –1.23 moles C 6 H 12 O 6 x 6 moles C Atoms_____ x 6.02 x Atoms C 1 mole C 6 H 12 O 6 Molecules 1 mole C atoms 1 mole C 6 H 12 O 6 Molecules 1 mole C atoms = 4.44 x atoms C

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10 Measuring Moles n The decimal number on the periodic table is –The mass of the average atom in amu –The mass of 1 mole of those atoms in grams.

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11 Gram Atomic Mass n Gram Atomic Mass is the mass of 1 mole of an element in grams. n We can write this as g C = 1 mole n We can count things by weighing them.

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12 Examples n How much would 2.34 moles of carbon weigh? 2.34 moles C x 12 g C__ = 28.1 g C 1 mole C 1 mole C

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13 Examples n How many moles of magnesium in 4.61 g of Mg? 4.61 g Mg x 1 mole Mg = 0.19 moles Mg 4.61 g Mg x 1 mole Mg = 0.19 moles Mg 24.3 g Mg 24.3 g Mg

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14 Examples n How many atoms of lithium in 1.00 g of Li? 1.00 g Li x 1 mole Li x 6.02 x atoms Li = 8.7 x atoms Li 6.9 g Li 1 mole Li 6.9 g Li 1 mole Li

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15 Examples n How much would 3.45 x atoms of U weigh? 3.45 x atoms U x 1 mole U_________ x 238g U = 13.6 g U 3.45 x atoms U x 1 mole U_________ x 238g U = 13.6 g U 6.02 x atoms U 1 mole U 6.02 x atoms U 1 mole U

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16 What about compounds? n In 1 mole of H 2 O molecules there are two moles of H atoms and 1 mole of O atoms. (Be careful) n To find the mass of one mole of a compound –Determine the moles of the elements contained in the compound. –Find out how much each would weigh –add them up

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17 What about compounds? n Molar Mass is the mass of one mole of any compound. Example: What is the Molar mass of CH 4 ? »1 mole of C = g »4 mole of H = 4 x 1.01 g = 4.04g »1 mole CH 4 = = 16.05g/mol

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18 Molar Mass n What is the molar mass of Fe 2 O 3 ? »2 moles of Fe = 2 x g = g »3 moles of O = 3 x g = g »The Molar mass = g g = g/mol

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19 Calculate the molar mass of the following n C 6 H 12 O 6 »C = 6 x 12g = 72g »H = 12 x 1.0g = 12g »O = 6 x 16g = 96g n Molar mass = = 180g/mol n (NH 4 ) 3 PO 4 »N = 3 x 14g = 42g »H = 12 x 1.0g = 12g »P = 1 x 30.9g = 30.9g »O = 4 x 16g = 64g n Molar mass = 42g + 12g g + 64g = 148.9g/mol

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20 Using Molar Mass Finding moles of compounds Counting pieces by weighing

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21 Molar Mass n The number of grams in 1 mole of atoms, formula units, or molecules. n We can make conversion factors from these. n To change grams of a compound to moles of a compound. n Or moles to grams

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22 For example n How many moles is 5.69 g of NaOH? –5.69 g NaOH x 1 mole NaOH Molar mass of NaOH Molar mass of NaOH Molar Mass NaOH = Na + O + H = 23.0g+16g + 1.0g = 40g/mole Substitute 40g/mole into the equation above and get: –5.69 g NaOH x 1 mole NaOH = 0.14 moles NaOH 40 g NaOH 40 g NaOH

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23 Gases and the Mole

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24 Gases n Many of the chemicals we deal with are gases. n They are difficult to weigh, so well measure volume n Need to know how many moles of gas we have. n Two things affect the volume of a gas –Temperature and pressure –Compare at the same temp. and pressure.

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25 Standard Temperature and Pressure n Avogadro's Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles. n 0ºC and 1 atmosphere pressure n Abbreviated atm n 273 K and kPa n kPa is kiloPascal

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26 At Standard Temperature and Pressure n abbreviated STP n At STP 1 mole of gas occupies 22.4 L n Called the molar volume n Used for conversion factors n Moles to Liter (x 22.4), and L to mol (divide by 22.4)

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27 Examples n What is the volume of 4.59 mole of CO 2 gas at STP? 4.59 mole CO 2 x 22.4L CO 2 = L CO 2 1 mole CO 2 1 mole CO 2

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28 Density of a gas n D = m /V n for a gas the units will be g / L n We can determine the density of any gas at STP if we know its formula. n To find the density we need the mass and the volume. n If you assume you have 1 mole than the mass is the molar mass (PT) n At STP the volume is 22.4 L.

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29 Examples n Find the density of CO 2 at STP. Note: At STP the volume is 22.4 L D =m/v = Molar mass CO 2 = ((12g + (16g x 2g)) = 1.96g/l 22.4 L 22.4L 22.4 L 22.4L

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30 Quizdom n Find the density of CH 4 at STP. D=m/v ((12.0g +(1.0g x 4)) = 2.14 g/l 22.4L 22.4L

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31 The other way n Given the density, we can find the molar mass of the gas. n Again, pretend you have a mole at STP, so V = 22.4 L. n m = D x V n m is the mass of 1 mole, since you have 22.4 L of the stuff. n What is the molar mass of a gas with a density of g/L? M=D x V = 1.964g/l x 22.4L = 44g

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32 All the things we can change

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33 Volume Ions Atoms Representative Particles Mass PT Moles 6.02 x L Count

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34 Percent Composition n Like all percents n Part x 100 % whole n Find the mass of each component, n divide by the total mass.

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35 Example n Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S. Part x 100 Whole –Total weight = 29.0g Ag g S = 33.3 g %Ag = 29.0g Ag x 100 = 87.1% Ag 33.3 g 33.3 g % S = 4.30g S x 100 = 12.9% S 33.3g 33.3g

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36 Getting it from the formula n If we know the formula, assume you have 1 mole. n Then you know the pieces and the whole.

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37 Examples n Calculate the percent composition of C 2 H 4 ? –Assume 1 mole –1 mole C 2 H 4 = (12g x 2) +(1.0g x 4) = 28g/mole total mass % C = (12g x 2) x 100 = 85.7% C % C = (12g x 2) x 100 = 85.7% C 28 g 28 g % H = (1.0g x 4) x 100 = 14.3% H 28 g 28 g

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38 Examples n What is the percent composition of Aluminum carbonate. –Write the formula: Al 2 (CO) 3 (ionic Formula writing) –Assume 1 mole –Calculate Total Mass: (Al x 2) + (C x 3) + (O x 3) = (27g x 2) + (12g x 3) + (16g x 3) = 54g + 36g + 48g = 138g »% Al = (27g x 2) x 100 = 39.1%Al 138g 138g % C = (12g x 3) x 100 = 26.1% C % C = (12g x 3) x 100 = 26.1% C 138g 138g % O = (16g x 3) x 100 = 34.8% O % O = (16g x 3) x 100 = 34.8% O 138g 138g

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39 Percent to Mass n Multiply % by the total mass to find the mass of that component. n How much aluminum in 450 g of aluminum carbonate? – Write the formula: Al 2 (CO) 3 (ionic Formula writing) »% Al = (27g x 2) x 100 = 39.1%Al 138g 138g % C = (12g x 3) x 100 = 26.1% C % C = (12g x 3) x 100 = 26.1% C 138g 138g % O = (16g x 3) x 100 = 34.8% O % O = (16g x 3) x 100 = 34.8% O 138g 138g –39.1%Al x1/100 x 450g Al 2 (CO) 3 = 176g Al

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40 Empirical Formula From percentage to formula

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41 The Empirical Formula n The lowest whole number ratio of elements in a compound. n The molecular formula the actual ratio of elements in a compound. n The two can be the same. n CH 2 empirical formula n C 2 H 4 molecular formula n C 3 H 6 molecular formula n H 2 O both

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42 Finding Empirical Formulas n Just find the lowest whole number ratio n C 6 H 12 O 6 n CH 4 N 2 n It is not just the ratio of atoms, it is also the ratio of moles of atoms.

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43 Calculating Empirical Formulas n Means we can get ratio from percent composition. n Assume you have a 100 g. n The percentages become grams. n Turn grams to moles. n Find lowest whole number ratio by dividing everything by the smallest moles.

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44 Example n Calculate the empirical formula of a compound composed of % C, % H, and %N. n Assume 100 g so n g C x 1mol C = mole C gC n g H x 1mol H = 16.1 mole H 1.01 gH n g N x 1mol N = mole N gN

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45 Example n The ratio is mol C = 1 mol C molN 1 mol N n The ratio is 16.1 mol H = 5 mol H molN 1 mol N nC1H5N1nC1H5N1nC1H5N1nC1H5N1 n Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

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46 Empirical to molecular n Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? n Since the empirical formula is the lowest ratio the actual molecule would weigh the same or more. n By a whole number multiple. n Divide the actual molar mass by the the mass of one mole of the empirical formula. n You will get a whole number. n Multiply the empirical formula by this.

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47 Example n n A compound has an empirical formula of ClCH 2 and a molar mass of g/mol. What is its molecular formula? n n A compound has an empirical formula of CH 2 O and a molar mass of g/mol. What is its molecular formula?

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48 Percent to molecular n Take the percent x the molar mass –This gives you mass in one mole of the compound n Change this to moles –You will get whole numbers –These are the subscripts n Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. It has a molar mass of 194 g. What is its molecular formula?

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49 Example n n Ibuprofen is % C, 8.80 % H, % O, and has a molar mass of about 207 g/mol. What is its molecular formula?

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