MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY

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1 MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY [MH5; Ch. 3] Each element has its own unique mass. The mass of each element is found on the Periodic Table under the chemical symbol for the element (it is usually given to several decimal places); this may be referred to as the ATOMIC MASS (or ATOMIC WEIGHT). Atomic masses are relative measurements; based on how heavy an element is compared to another element. The mass of 1 atom of an element may be expressed in amu, short for Atomic Mass Units. The modern scale of atomic masses is based on the most common isotope of carbon; 12 C... This isotope is assigned a mass of 12 amu: Mass of one 12 C atom = 12 amu (exactly) So... 1 amu = 1 x mass of one C atom 12 6 When determining atomic weight, we take a large number of atoms with isotopes present in their natural abundances. Atomic weights of all elements except 12 C are subject to experimental error; ( 12 C has atomic mass of exactly 12 amu). For elements that are isotopically pure (consist of one isotope only; monoisotopic), we may talk of atomic mass (mass of one atom) 12 17

2 EXAMPLE: Sodium, Na (only natural isotope), At. Wt = So the mass of 1 atom of Na is amu. Notice that the atomic mass of the element is very close to its mass number... For elements that naturally contain more than one isotope, we can determine an average atomic mass; an average of the isotopic masses weighted according to their fractional relative natural abundances. This average is numerically the same as atomic weight... EXAMPLE 1: Chlorine 35 CR isotopic mass: amu, 75.77% 37 CR isotopic mass: amu, 24.23% (Note: these are atom % s, not mass % s) What is the average atomic mass? 18

3 EXAMPLE 2: Antimony, Sb, consists of two isotopes; 121 Sb (atomic mass ) and 123 Sb (atomic mass ) Calculate the relative abundances (in atom %) of these two isotopes. The Average Atomic Mass for any element is found on the Periodic Table. 19

4 How do we calculate the mass of a molecule? This is called Molecular Weight (or Mass): it is the sum of the atomic weights of all the atoms in the molecule. EXAMPLE 1: O 2 EXAMPLE 2: H 2 O EXAMPLE 3: Mg(NO 3 ) 2 We never actually measure the mass of one atom or one molecule of anything...an amu is an extremely small measurement. Chemists like to measure masses in grams......so what is the relationship between amu and grams? Before we can answer this question, we must get a very clear idea of what a mole is... 20

5 Demystifying The Mole A mole is a number; a very large number. It is a word we use to represent a certain number of things. We use it because it is easier to say the word than to say the number itself... We have commonly used names for many groups of items... Quantity Number of Items pair trio quartet dozen gross mole So, the mole (which we abbreviate as mol) contains Avogadro s Number (N A ) of items: N A = 6.02 x This is a really strange number, but we use it because of the official definition of the mole; which is the number of atoms in exactly 12 g of the isotope 12 C. It turns out that this number of atoms is Avogadro s Number, or 6.02 x Therefore, 1 mole of anything contains 6.02 x of those things. 12 g of 12 C is one mole of 12 C and contains N A (Avogadro s No.) atoms (by definition). 21

6 Atoms and molecules are really small, and in any sample of measurable size (usually in g or mg) we have a lot of them; so we use the term moles as a sort of shorthand way to express how many molecules or atoms we are actually talking about. Back to the question: Relating the amu to grams g of 12 C is one mole of 12 C and contains N A (Avogadro s Number) of atoms... This is another very small number...and remember that we normally do not measure the mass of just one atom or molecule. It is much more practical to measure the mass of a larger amount of a substance; so we use the mass of one mole of the substance... The MOLAR MASS is the mass in grams of 1 mole of compound. EXAMPLES: O 2 : H 2 O: Mg(NO 3 ) 2 : Notice that the number (from the Periodic Table) is the same as the Molecular Weight; it is the units that are different!! 22

7 One atom (or molecule) of substance will have a mass given in amu. One mole of atoms (or molecules) of a substance will have a mass given in grams. (Molar Mass is usually given units of g mol 1 ) Some compounds exist as "hydrated" molecules; for each molecule of compound, a certain number of water molecules are associated with that molecule. Think of it as a molecule of compound being surrounded by a certain number of water molecules. To determine the Molar Mass of one of these compounds, the mass of the water molecules associated with the molecule must be added to the mass of the molecule. EXAMPLE: The Molar Mass of CuSO 4 C 5 H 2 O We use Molar Mass to convert from moles to g or vice-versa. 23

8 EXAMPLE 1: a) How many moles of CO 2 molecules are there in a 55.0 g sample of CO 2? b) How many moles of CO 2 molecules are there in a 25.0 g sample of CO 2? 24

9 EXAMPLE 2: What is the mass of moles of C 2 H 5 OH? EXAMPLE 3: Suppose we have a sample of a compound with a mass of 2.68 g. An experiment has shown that in this sample there are 1.26 x 10 2 moles of the compound in this sample. What is the molar mass of the compound? 25

10 CHEMICAL FORMULAE Recall that a chemical formula is the ratio (in whole numbers) of the atoms in a compound... EXAMPLE: CH 4 (Methane) 1 molecule of CH 4 contains: 1 mole of CH 4 molecules contains: How many atoms would this be?? 26

11 PERCENTAGE COMPOSITION The composition of a compound is often expressed by mass as % s, because that is the way elemental analyses are reported. % composition by mass is found by using atomic masses and the formula of the compound. EXAMPLE: Using C 8 H 18, find the % composition; i) by mass ii) by moles i) C 8 H 18 has a Molar Mass of C 8 H 18 contains 8 C (at g mol -1 each) and 18 H (at g mol -1 ) % C (by mass) = % H (by mass) = We could use this method to determine the MASS FRACTION of one of the elements in a compound. In reporting the composition of C 8 H 18 by mass fractions, we would say that the fraction of C 8 H 18 that is Carbon is: and the fraction that is Hydrogen is: 27

12 ii) Mole percent uses the same principle, but is reporting the composition using moles instead of mass. In one mole of C 8 H 18 molecules, there are 8 moles of Carbon atoms and 18 moles of Hydrogen atoms. The total number of moles of atoms is twenty-six (26). The mole % of Carbon = The mole % of Hydrogen = The term MOLE FRACTION (denoted by X) is used frequently... The mole fraction is the mole percent reported as a fraction. We could say that: The mole fraction of Carbon in C 8 H 18 is: The mole fraction of Hydrogen is: 28

13 Can we deduce a formula from % composition? Yes!! EXAMPLE: A compound contains 80.1 % C and 19.9 % H by mass. Find the SIMPLEST FORMULA of the compound. (The simplest, or empirical formula shows the simplest ratio of atoms in the compound.) Given % by mass, we use a 100 g sample of compound; therefore, we have 80.1 g of C and 19.9 g of H. Find the number of moles of each element first... To find the simplest formula, we want a whole number ratio, so divide each # moles by the smallest # moles... 29

14 The MOLECULAR FORMULA of a compound is a multiple of the simplest formula; for CH 3, we could write the molecular formula as (CH 3 ) n. To find "n", we need the Molar Mass of the compound's molecular formula... EXAMPLE: If we know that the Molar Mass of (CH 3 ) n is 30 g mol -1, we can say: 30

15 Sometimes you will not be given the elemental analysis, only some experimental data. From the experimental data you will need to determine the amounts of each element present in the compound. EXAMPLE: A compound contains only C, H and O. When burned in excess O 2, a g sample of compound yields g CO 2 and g H 2 O. If the Molar Mass of the compound is 90 ± 5 g mol -1, find the molecular formula. To find molecular formula, we must first find simplest formula. We need either % composition, or the masses of each element in the compound. This will enable us to find the # moles of each element, and therefore the mole ratio ( the simplest formula). For C : 31

16 For H : For O: 32

17 Now find the formula... 33

18 Sometimes we do not know the identity of all the elements in a compound, but we do know the percentage of one element in said compound... We can use this percentage to determine the molar mass, and then the identity of the unknown element. EXAMPLE: An oxide of an unknown metal has the formula X 3 O 4. Analysis shows that this compound is 27.64% oxygen by mass. Determine the molar mass and identity of X. 34

19 MASS RELATIONS IN REACTIONS: [MH5; 3.4] Stoichiometry ; a term used to describe the relationship between amounts of reactants and products (in moles and grams) using balanced chemical equations. In a chemical equation, symbols of elements or compounds represent MOLES... 2 S + 3 O 2! 2 SO 3 two moles three moles two moles of of Sulfur of Oxygen Sulfur Trioxide ----REACTANTS PRODUCTS---- What is this equation telling us? 1) 2 moles of Sulfur will react with 3 moles of Oxygen. 2) 2 moles of Sulfur will produce 2 moles of Sulfur Trioxide. 3) 3 moles of Oxygen will produce 2 moles of Sulfur Trioxide. Notes: a) Equations must be balanced in mass of each element, and in charge. b) By convention, coefficients in an equation are usually integers, the smallest values possible to balance. S + O 2! SO 3 S + O 2! SO 3 S + O 2! SO 3 35

20 c) Use a one-way arrow (!) for a reaction going mainly forwards : C + O 2! CO 2 Use a two-way arrow ( º ) for an equilibrium: 2NO + O 2 º 2NO 2 (In an equilibrium situation, the reaction is proceeding in both directions.) d) Indicate the phase of reactants and products, if required, by the symbols: (s) solid (R) liquid (g) gas (aq) aqueous sol n 2 Na (s) + 2 H 2 O (R)! 2 Na + (aq) + 2 OH - (aq) + H 2 (g) e) Remember the difference between atoms and molecules. Don t write a symbol for an atom when an element occurs as a diatomic molecule: H 2 (g) N 2 (g) O 2 (g) F 2 (g) CR 2 (g) Br 2 (R) I 2 (s) Balancing Equations - this is essential! Some are easy: C(s) + O 2 (g)! CO 2 (g) Others need work: N 2 + H 2! NH 3 36

21 f) Combustion Reactions are important - what happens when octane (a component of gasoline) burns in oxygen? C 8 H 18 + O 2! CO 2 + H 2 O C 8 H 18 + O 2! CO 2 + H 2 O Always check your equation AFTER balancing it!! g) By knowing the ratios by which species react to form products, we can determine the number of moles of species formed or reacted, given the number of moles of another species present in the reaction. EXAMPLE 1: If 2.00 g of C 2 H 4 are burnt in excess O 2, how many grams of CO 2 are formed? First, balance the equation: C 2 H 4 + O 2! CO 2 + H 2 O The steps to solving this problem are: 1) Find number of moles of C 2 H 4. 2) Find number of moles of CO 2 by relating moles reactant (which we know) to moles product (which is what we want). 3) Find mass CO 2. 37

22 The solution is: 1) moles C 2 H 4 2) Relate # moles of reactant to # moles product. (from equation) 3) Mass CO 2 = EXAMPLE 2: White phosphorous, P 4, may be burned in oxygen, O 2, to produce P 4 O 10. What mass of oxygen is needed to completely burn (or react with) 1.20 g of white phosphorus? First, write the balanced equation: P 4 + O 2! P 4 O 10 Then find the number of moles of P 4 : (MM = 4 x gmol 1 ) 38

23 Write what is happening in words: Set up a ratio and solve it. Finish the calculation... Note: The Molar Mass you use must match the species in the equation! LIMITING REAGENT problems involve reactions where the amount of product formed is limited by the amount of reagent available to react. EXAMPLE 1: You are assembling bicycles. You have 6 frames and 10 wheels. How many bicycles can you build? 39

24 EXAMPLE 2: What mass of AR 2 (SO 4 ) 3 can be prepared by reacting 25.0 g of AR with 100 g of H 2 SO 4? The reaction is: AR + H 2 SO 4! AR 2 (SO 4 ) 3 + H 2 First things first: Make sure that the equation is balanced!!! The clue that this is a limiting reagent problem is that the amounts of both starting materials are given... These amounts may be given in various ways; they will all enable you to find the # moles of each reactant. Find # moles of each reactant: TO FIND THE LIMITING REAGENT: Divide the actual number of moles of each reactant by its coefficient (the required number of moles) from the balanced equation; then compare the resultant numbers. The smallest of these two numbers tells you the limiting reagent! 40

25 It is important to realize that you only use this step to determine the limiting reagent; you do not necessarily use these numbers again (although you often see them again!) You can only form as much product as you have limiting reagent available to react...so all calculations must relate the amount of limiting reagent to another species. Relate # moles of limiting reagent to # moles product... Set up a ratio: Finish the calculation: 41

26 YIELD CALCULATIONS Yield is usually expressed on a % basis, showing the amount of product obtained as a % of what would be expected from the equation: moles obtained x 100 % OR mass obtained x 100 % moles expected mass expected EXAMPLE 1: When 12.0 g iron, Fe, were burned in air, 13.8 g iron oxide, Fe 3 O 4, were produced. Calculate the % yield of Fe 3 O 4 in this reaction. For % Yield, remember... moles obtained x 100 % = mass obtained x 100% moles expected mass expected Write the balanced equation: Fe + O 2! Fe 3 O 4 Find number of moles of Fe: 42

27 From the equation, we would expect to obtain: We actually obtained 13.8 g; less than we would expect... The PERCENTAGE YIELD would be: EXAMPLE 2: A 5.00 g sample of S was chemically treated and several steps later resulted in the retrieval of 41.6 g of Na 2 SO 4 C10 H 2 O. Calculate the % yield. All the intermediate steps may be ignored! Since the final product contains 1 S atom; one mole S gives one mole of Na 2 SO 4 C10H 2 O (this statement serves as the balanced equation!!) 43

28 As usual, start by determining the number of moles... The molar mass of Na 2 SO 4 C10H 2 O is: We would expect a yield of: Mass of Na 2 SO 4 C10H 2 O expected: 44

29 EXAMPLE 3: We have seen that in most chemical reactions, one of the reactants is present in a lesser amount; this is called the Limiting Reagent. Some reactions take more than one step to occur; we saw that the moles of the various species carry all the way through the entire process. Here s an example which incorporates almost all the intricacies involved with stoichiometry. EXAMPLE: Copper metal is usually obtained from sulfide ores, such as Cu 2 S. The refining process takes two steps; the first is roasting the Cu 2 S in oxygen, O 2, to obtain copper (I) oxide, Cu 2 O. In the second step, the Cu 2 O is reacted with powdered carbon, a process which yields carbon monoxide gas and copper metal. 1) 2 Cu 2 S (s) + 3 O 2 (g)! 2 Cu 2 O (s) + 2 SO 2 (g) 2) Cu 2 O (s) + C (s)! 2 Cu (s) + CO (g) A) What is the overall equation? -45-

30 B) Suppose we react 10.0 grams of Cu 2 S with 4.50 grams of O 2. What yield of Cu 2 O would be expected in step 1)? C) If 5.16 g of Cu (s) results from the second step, what is the percentage yield from the entire process? -46-

31 % PURITY Some problems involve the use of reactants that are not 100 % pure. EXAMPLE 1: A solution of CS 2 (R) has a purity of 76.5 % (by mass) and has a density of 1.20 gml 1. What mass of SO 2 (g) could be produced if 15.0 ml of CS 2 were burned in excess oxygen? 47

32 EXAMPLE 2: An ore contains 35.0 % by mass iron, Fe. What mass of this ore is needed to produce 500 g of Fe 2 O 3? Write a balanced equation: Fe + O 2! Fe 2 O 3 Find the number of moles of whatever you can... Moles Fe 2 O 3 = From the equation, we know: 48

33 Mass of Fe required: Set up a ratio to relate the % purity of the ore to the amount of Fe that is needed: 49

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