Mass Relationships in Chemical Reactions

Mass Relationships in Chemical Reactions
Chapter 3

Mass and Moles of a Substance
Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows the chemist to carry out “recipes” for compounds based on the relative numbers of atoms involved. The calculation involving the quantities of reactants and products in a chemical equation is called stoichiometry. 2

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 = amu 16O = amu

Molecular Weight and Formula Weight
The molecular weight of a substance is the sum of the atomic weights of all the atoms in a molecule of the substance. For, example, a molecule of H20 contains 2 hydrogen atoms (at 1.0 amu each) and 1 oxygen atom (16.0 amu), giving a molecular weight of 18.0 amu. 2

Average atomic mass of lithium:
Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu) Average atomic mass of lithium: 7.42 x x 7.016 100 = amu

Pg 78b

Average atomic mass (6.941)

The Mole Concept A mole is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon–12. The number of atoms in a 12-gram sample of carbon–12 is called Avogadro’s number (to which we give the symbol Na). The value of Avogadro’s number is 6.02 x 1023. 2

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 grams of 12C 1 mol = NA = x 1023 Avogadro’s number (NA)

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 = x 1023 atoms = g 1 12C atom = amu 1 mole 12C atoms = g 12C 1 mole lithium atoms = g of Li For any element atomic mass (amu) = molar mass (grams)

One Mole of: S C Hg Cu Fe

The molar mass of a substance is the mass of one mole of a substance. For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units. That is, one mole of any element weighs its atomic mass in grams. 2

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

Mole calculations Suppose we have grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent? 2

Mole calculations Conversely, suppose we have 5.75 moles of magnesium (atomic wt. = 24.3 g/mol). What is its mass? 2

Do You Understand Molar Mass?
How many atoms are in g of potassium (K) ? 1 mol K = g K 1 mol K = 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 Na. 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.

Molecular Weight and Formula Weight
The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not. For example, one formula unit of NaCl contains 1 sodium atom (23.0 amu) and one chlorine atom (35.5 amu), giving a formula weight of 58.5 amu. 2

formula mass (amu) = molar mass (grams)
Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. NaCl 1Na 22.99 amu 1Cl amu NaCl 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = amu 1 mole NaCl = 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 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 amu SO2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = amu 1 mole SO2 = g SO2

Mole calculations This same method applies to compounds. Suppose we have grams of H2O (molecular weight = 18.0 g/mol). How many moles does this represent? 2

Mole calculations Conversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = g/mol). 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 = 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% % % = 100.0%

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 percent. nK = g K x = mol K 1 mol K 39.10 g K nMn = g Mn x = mol Mn 1 mol Mn 54.94 g Mn nO = g O x = mol O 1 mol O 16.00 g O

Percent Composition and Empirical Formulas
nK = , nMn = , 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 Divide by smallest subscript (0.25) Empirical formula C2H6O

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 the empirical formula from the percent composition. Our 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 the empirical formula from the percent composition. Our 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 the “true” molecular formula from the empirical formula. An empirical formula gives only the smallest whole-number ratio of atoms in a formula. The “true” molecular formula could be a multiple of the empirical formula (since both would have the same percent composition). To determine the “true” molecular formula, we must know the “true” molecular weight of the compound. 2

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/mol. The molar weight of the empirical formula (the “empirical weight”) is only 30.0 g/mol. This would imply that the “true” molecular formula is actually the empirical formula doubled, or C2H4O2 2

3 ways of representing the reaction of H2 with O2 to form H2O
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 + O 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 grams O2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O2 makes 2 g MgO

Features of a Chemical Equation
 - means heat is needed. Products and reactants must be specified using chemical symbols Reactants- written on the left of arrow Products - written on the right Physical states are shown in parentheses

All the atoms of every reactant must also appear in the products.
Coefficient - how many of the substance are in the reaction The equation must be balanced. All the atoms of every reactant must also appear in the products. How many Hg’s on left? on right? How many O’s on left?

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.

Acid and base are mixed together
Heat is produced or absorbed (temperature changes) Acid and base are mixed together The color changes Light is absorbed or emitted Changes in the way the substances behave in an electrical or magnetic field Changes in electrical properties.

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

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

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

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

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

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 C2H O2 2CO2 + 3H2O 7 2 remove fraction multiply both sides by 2 2C2H6 + 7O2 4CO2 + 6H2O

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

Balance the following equations. 2

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

This balanced chemical equation shows that one mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. 1 molecule N molecules H 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.

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

Mass Relationships in Chemical Equations
Amounts of substances in a chemical reaction by mass. How many grams of HCl are required to react with 5.00 grams manganese dioxide according to this equation? 2

Mass Relationships in Chemical Equations
First, you write what is given (5.00 g MnO2) and convert this to moles. Then convert to moles of what is desired.(mol HCl) Finally, you convert this to mass (g HCl) 2

Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH3OH moles CH3OH moles H2O grams H2O molar mass CH3OH coefficients chemical equation molar mass H2O 1 mol CH3OH 32.0 g CH3OH x 4 mol H2O 2 mol CH3OH x 18.0 g H2O 1 mol H2O x = 209 g CH3OH 235 g H2O

Let's do some examples. Na + Cl2  NaCl 1. Balance the equation.
2. Calculate the moles of Cl2 which will react with 5.00 mol Na. 3. Calculate the number of grams of NaCl which will be produced when 5.00 mol Na reacts with an excess of Cl2. 4. Calculate the grams of Na which will react with 5.00 g Cl2.

Limiting Reagent The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. The limiting reagent ultimately determines how much product can be obtained. For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, it is clear that the number of frames will determine how many bicycles can be made. 2

Limiting Reagents 2NO + 2O2 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

Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O Al2O3 + 2Fe Calculate the mass of Al2O3 formed. g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Al 27.0 g Al x 1 mol Fe2O3 2 mol Al x 160. g Fe2O3 1 mol Fe2O3 x = 124 g Al 367 g Fe2O3 Start with 124 g Al need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent

Use limiting reagent (Al) to calculate amount of product that
can be formed. g Al mol Al mol Al2O3 g Al2O3 2Al + Fe2O Al2O3 + 2Fe 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al x 102. g Al2O3 1 mol Al2O3 x = 124 g Al 234 g Al2O3

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

Operational Skills Calculating the formula weight from a formula.
Calculating the mass of an atom or molecule. Converting moles of substance to grams and vice versa. Calculating the number of molecules in a given mass. Calculating the percentage composition from the formula. Calculating the mass of an element in a given mass of compound. Calculating the percentages C and H by combustion. Determining the empirical formula from percentage composition. Determining the true molecular formula. Relating quantities in a chemical equation. Calculating with a limiting reagent. 2

Worked Example 3.4

Worked Example 3.6

Worked Example 3.7

Worked Example 3.9

Worked Example 3.11

Worked Example 3.12

Worked Example 3.13a

Worked Example 3.13b

Worked Example 3.15a

Mass Relationships in Chemical Reactions

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