Mass Relationships in Chemical Reactions Chapter 3 Mass Relationships in Chemical Reactions

Atomic mass is the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu 16O = 16.00 amu

Copper, a metal known since ancient times, is used in electrical cables and pennies, among other things. The atomic masses of its two stable isotopes, (69.09 percent) and (30.91 percent), are 62.93 amu and 64.9278 amu, respectively. Calculate the average atomic mass of copper. The relative abundances are given in parentheses. Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of that particular isotope. (0.6909) (62.93 amu) + (0.3091) (64.9278 amu) = 63.55 amu

Avogadro’s number (NA) The Mole (mol): A unit to count numbers of particles Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = NA = 6.0221415 x 1023 Avogadro’s number (NA)

Avogadro’s number (NA) 6.022 x 1023 = NA = 1mol

Molar mass is the mass of 1 mole of in grams marbles atoms eggs shoes Molar mass is the mass of 1 mole of in grams marbles atoms 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams)

One Mole of: S C Hg Cu Fe

1 12C atom 12.00 amu 12.00 g 6.022 x 1023 12C atoms = 1.66 x 10-24 g 1 amu x M = molar mass in g/mol NA = Avogadro’s number

3.2 Helium (He) is a valuable gas used in industry, low-temperature research, deep-sea diving tanks, and balloons. How many moles of He atoms are in 6.46 g of He? Thus, there are 1.61 moles of He atoms in 6.46 g of He. Check Because the given mass (6.46 g) is larger than the molar mass of He, we expect to have more than 1 mole of He.

Moles to g How many grams of gold (Au) are in 15.3 moles of Au? 1mole of Au = 197.0 g 15.3 moles of Au = (197.0 g/1mol)x 15.3 mol = 3014.0 g =3.01x103 g

Mass and Moles of a Substance Mole calculations Suppose we have 100.0 grams of iron (Fe). The atomic mass of iron is 55.8 amu. How many moles of iron does this represent? 2

Mass and Moles of a Substance Mole calculations Conversely, suppose we have 5.75 moles of magnesium (molar mass = 24.3 g/mol). What is its mass? 2

Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 1 mol K 39.10 g K x x 6.022 x 1023 atoms K 1 mol K = 0.551 g K 8.49 x 1021 atoms K

Let's Practice Using these Tools 1. What is the mass in grams of 2.5 mol Ca 2. Calculate the number of atoms in 1.7 mol B. 3. Calculate the number of atoms in 5.0 g Al. 4. Calculate the mass of 5,000,000 atoms of Au.

formula mass (amu) = molar mass (grams) Molecular Weight and Formula Weight Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. 1Na 22.99 amu 1Cl + 35.45 amu NaCl 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl

Do You Understand Formula Mass? What is the formula mass of Ca3(PO4)2 ? 1 formula unit of Ca3(PO4)2 3 Ca 3 x 40.08 2 P 2 x 30.97 8 O + 8 x 16.00 310.18 amu

molecular mass (amu) = molar mass (grams) Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO2 1S 32.07 amu 2O + 2 x 16.00 amu SO2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2

Mass and Moles of a Substance Mole calculations This same method applies to compounds. Suppose we have 100.0 grams of H2O (molar mass = 18.0 g/mol). How many moles does this represent? 2

Mass and Moles of a Substance Mole calculations Conversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = 180.0 g/mol). (the correct term is molar mass, but this is a common usage) What is its mass? 2

Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 1 mol C3H8O 60 g C3H8O x 8 mol H atoms 1 mol C3H8O x 6.022 x 1023 H atoms 1 mol H atoms x = 72.5 g C3H8O 5.82 x 1024 atoms H

Heavy Light

Determining Chemical Formulas The percent composition of a compound is the mass percentage of each element in the compound. We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is, 2

Mass Percentages from Formulas Let’s calculate the percent composition of butane, C4H10. First, we need the molecular mass of C4H10. Now, we can calculate the percents. 2

Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% C2H6O %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0%

Determining Chemical Formulas Determining the formula of a compound from the percent composition. The percent composition of a compound leads directly to its empirical formula. An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts. 2

Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. nK = 24.75 g K x = 0.6330 mol K 1 mol K 39.10 g K nMn = 34.77 g Mn x = 0.6329 mol Mn 1 mol Mn 54.94 g Mn nO = 40.51 g O x = 2.532 mol O 1 mol O 16.00 g O

Percent Composition and Empirical Formulas nK = 0.6330, nMn = 0.6329, nO = 2.532 K : ~ 1.0 0.6330 0.6329 Mn : 0.6329 = 1.0 O : ~ 4.0 2.532 0.6329 KMnO4

g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O g CO2 mol CO2 mol C g C 6.0 g C = 0.5 mol C g H2O mol H2O mol H g H 1.5 g H = 1.5 mol H g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Empirical formula C0.5H1.5O0.25 See p. 88-89 Divide by smallest subscript (0.25) Empirical formula C2H6O

Determining Chemical Formulas Determining the empirical formula from the percent composition. Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula? In other words, give the smallest whole-number ratio of the subscripts in the formula Cx HyOz 2

Determining Chemical Formulas Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain: This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might emerge. 2

Determining Chemical Formulas Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain: now it’s not too difficult to see that the smallest whole number ratio is 7:6:2. The empirical formula is C7H6O2 . 2

Determining Chemical Formulas Determining the “true” molecular formula from the empirical formula. For example, suppose the empirical formula of a compound is CH2O and its “true” molecular weight is 60.0 g. The molar weight of the empirical formula (the “empirical weight”) is only 30.0 g. This would imply that the “true” molecular formula is actually the empirical formula doubled (60g/30g = 2), or C2H4O2 2

3.11 A sample of a compound contains 30.46 percent nitrogen and 69.54 percent oxygen by mass, as determined by a mass spectrometer. In a separate experiment, the molar mass of the compound is found to be between 90 g and 95 g. Determine the molecular formula and the accurate molar mass of the compound.

3.11 Strategy To determine the molecular formula, we first need to determine the empirical formula. Comparing the empirical molar mass to the experimentally determined molar mass will reveal the relationship between the empirical formula and molecular formula. Solution We start by assuming that there are 100 g of the compound. Then each percentage can be converted directly to grams; that is, 30.46 g of N and 69.54 g of O.

3.11 Let n represent the number of moles of each element so that Thus, we arrive at the formula N2.174O4.346, which gives the identity and the ratios of atoms present. However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing the subscripts by the smaller subscript (2.174). After rounding off, we obtain NO2 as the empirical formula.

empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g 3.11 The molecular formula might be the same as the empirical formula or some integral multiple of it (for example, two, three, four, or more times the empirical formula). Comparing the ratio of the molar mass to the molar mass of the empirical formula will show the integral relationship between the empirical and molecular formulas. The molar mass of the empirical formula NO2 is empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g

3.11 Next, we determine the ratio between the molar mass and the empirical molar mass The molar mass is twice the empirical molar mass. This means that there are two NO2 units in each molecule of the compound, and the molecular formula is (NO2)2 or N2O4. The actual molar mass of the compound is two times the empirical molar mass, that is, 2(46.01 g) or 92.02 g, which is between 90 g and 95 g.

Chemical Reactions and Chemical Equations A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction 3 ways of representing the reaction of H2 with O2 to form H2O reactants products

How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O2 makes 2 g MgO

H2 is released when Na is placed in water. The Experimental Basis of a Chemical Equation We know that a chemical equation represents a chemical change. The following is evidence for a reaction: Release of a gas. CO2 is released when acid is placed in a solution containing CO32- ions. H2 is released when Na is placed in water. Formation of a solid (precipitate.) A solution containing Ag+ ions is mixed with a solution containing Cl- ions.

Chemical Reactions: Equations Writing chemical equations A chemical equation is the symbolic representation of a chemical reaction in terms of chemical formulas. For example, the burning of sodium and chlorine to produce sodium chloride is written The reactants are starting substances in a chemical reaction. The arrow means “yields.” The formulas on the right side of the arrow represent the products. 2

Chemical Reactions: Equations In many cases, it is useful to indicate the states of the substances in the equation. Writing chemical equations When you use these labels, the previous equation becomes 2

Chemical Reactions: Equations The law of conservation of mass dictates that the total number of atoms of each element on both sides of a chemical equation must match. The equation is then said to be balanced. Writing chemical equations Consider the combustion of methane to produce carbon dioxide and water. 2

Chemical Reactions: Equations Writing chemical equations For this equation to balance, two molecules of oxygen must be consumed for each molecule of methane, producing one molecule of CO2 and two molecules of water. Now the equation is “balanced.” 2 2

Balancing Chemical Equations Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C2H6 NOT C4H12

Balancing Chemical Equations Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not O 1 carbon on right 2 carbon on left multiply CO2 by 2 C2H6 + O2 2CO2 + H2O 6 hydrogen on left 2 hydrogen on right multiply H2O by 3 C2H6 + O2 2CO2 + 3H2O

Balancing Chemical Equations Balance those elements that appear in two or more reactants or products. multiply O2 by 7 2 C2H6 + O2 2CO2 + 3H2O 2 oxygen on left 4 oxygen (2x2) + 3 oxygen (3x1) = 7 oxygen on right C2H6 + O2 2CO2 + 3H2O 7 2 remove fraction multiply both sides by 2 2C2H6 + 7O2 4CO2 + 6H2O

Balancing Chemical Equations Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O 4 C (2 x 2) 4 C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) Reactants Products 4 C 12 H 14 O

Chemical Reactions: Equations Balance the following equations. 2

Chemical Reactions: Equations Balance the following equations. 2 6 6 2 9 3 4 2

Stoichiometry: Quantitative Relations in Chemical Reactions Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the balanced chemical equation and on the relationship between mass and moles. Such calculations are fundamental to most quantitative work in chemistry. 2

Amounts of Reactants and Products Write balanced chemical equation Convert quantities of known substances into moles Use coefficients in balanced equation to calculate the number of moles of the sought quantity Convert moles of sought quantity into desired units

Molar Interpretation of a Chemical Equation The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as “mole-to-mole” relationships. For example, the Haber process for producing ammonia involves the reaction of hydrogen and nitrogen. 2

Molar Interpretation of a Chemical Equation This balanced chemical equation shows that one mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. 1 molecule N2 + 3 molecules H2 2 molecules NH3 Because moles can be converted to mass, you can also give a mass interpretation of a chemical equation. 2

Calculations Using the Chemical Equation 7 We will learn in this section to calculate quantities of reactants and products in a chemical reaction. Need a balanced chemical equation for the reaction of interest. Keep in mind that the coefficients represent the number of moles of each substance in the equation.

Molar Interpretation of a Chemical Equation Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2. Because the coefficients in the balanced equation represent mole-to-mole ratios, the calculation is simple. 2

Limiting Reagent: Reactant used up first in the reaction. 2NO + O2 2NO2 NO is the limiting reagent O2 is the excess reagent

Limiting Reagent Zinc metal reacts with hydrochloric acid by the following reaction. If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced? 2

Limiting Reagent Take each reactant in turn and ask how much product would be obtained if each were totally consumed. The reactant that gives the smaller amount is the limiting reagent. Since HCl is the limiting reagent, the amount of H2 produced must be 0.26 mol. 2

Reaction Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100

Theoretical and Percent Yield The theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants. The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). 2

Theoretical and Percent Yield To illustrate the calculation of percentage yield, recall that the theoretical yield of H2 in the previous example was 0.26 mol (or 0.52 g) H2. If the actual yield of the reaction had been 0.22 g H2, then 2

Summary Atomic mass unit (amu) Avagadro’s number Percent composition by mass of a compound, Empirical formula, molecular formula Chemical Equations Stoichiometry and limiting reagents Theoretical yield, actual yied, and percent yield

Homework Practice all the examples worked out in the book 3.2,3.5,3.9,3.13,3.14,3.16,3.22,3.24,3.25,3.26,3.38,3.40,3.48,3.50,3.52,3.59,3.60,3.66,3.67, 3.68,3.89