Average Atomic Mass. A new unit called the atomic mass unit (amu) was developed to deal with the very small units of mass for particles like the atom.

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1 Average Atomic Mass Since atoms are so small and the mass of individual atoms is also very small, it is not useful to use the units of grams or kilogram. A new unit called the atomic mass unit (amu) was developed to deal with the very small units of mass for particles like the atom. 1 amu = 1.66 x g It has been found to be useful that instead of using absolute masses, it is best compare the relative masses of atoms using a reference isotope as a standard. The carbon 12 isotope has been set as this standard. The carbon 12 isotope has been assigned a mass of exactly 12 amu.

2 For example, the oxygen atom has a relative atomic mass of amu. The Mole Another way to measure the amount of a substance is to count the number of particles in that substance. When talking about the number atoms, the number of individual atoms in the sample is ridiculously large. Counting the individual atoms is NOT PRACTICAL. Just as a dozen represents 12, 1 MOLE represents 6.02 x particles of a substance. 6.02x10 1mole 23 or 1mole 6.02x10 23

3 This number (Avogadro s Number) is named in honor of the Italian scientist Amedo Avogadro di Quarenga from whose work the concept was based. How can we use Avogadro s Number in a calculation? Example #1 How many moles of magnesium make up 1.25 x atoms of magnesium? Answer: In order to solve this problem, we must realize that the conversion needed is 6.02x10 1mole 23 or 1mole 6.02x10 23 Which form do we use??? Remember the techniques that we used when we did conversions factors What you want goes on top.

4 1mole Mg 1.25 x atoms Mg x10 atoms Mg =.208 moles Mg Example #2 How many atoms of sulfur are present in 0.75 moles of sulfur? Answer: In order to solve this problem, we must again realize that the conversion needed is 6.02x10 1mole 23 or 1mole 6.02x10 23 Which form do we use??? x10 atoms S 0.75 moles S 1mole S = 4.5 x atoms S

5 Molar Mass One atom of aluminum has a mass of 4.48 x g. This mass is determined by adding up the masses of the protons, neutrons and electrons within the aluminum atom. What would be the mass of 1 mole of aluminum atoms (or 6.02 x atoms of aluminum)? 6.02 x atoms 4.48 x g 1 atoms = = g g is the mass of 6.02 x atoms of aluminum OR g is the mass of 1 mole of aluminum

6 THEREFORE, g is the mass of 1 mole of Aluminum More commonly stated, g/mole is the MOLAR MASS of aluminum. The Atomic Weights of the elements on the Periodic Table also represent the Molar Masses of every element. Let s find the molar masses of some atoms: Molar Mass O =? Molar Mass W =? Molar Mass Mg =? How do we use Molar Mass?

7 Example #3 How many grams of iron are present in 1.78 moles of iron? Answer: In order to solve this problem, we must realize that the conversion needed is the molar mass of iron: g 1mole or 1mole g Which form do we use??? 1.78 moles Fe g 1 mole Fe Fe = 99.4 g Fe Example #4 How many moles of calcium are present in 5.78 g of calcium?

8 Answer: In order to solve this problem, we must realize that the conversation needed is the molar mass of calcium: mole g or Which form do we use??? 1mole g 5.78 g Ca 1 mole Ca g Ca = moles Ca

9 How can we determine the molar mass of a compound? 1) You must know the chemical formula of the compound. 2) Add the molar masses of the individual atoms in the molecule. Example: SO 3 The molar mass of sulfur is g/mole. The molar mass of oxygen is g/mole. Therefore, molar mass of SO 3 = 1(32.06 g/mole) + 3(16.00 g/mole) g/mole) Try this: What is the molar mass of BaBr 2?

10 Mole Mass Conversions The molar mass of an element or compound can be used to the convert mass of a substance into moles. Example #1: How many moles are in 6.59 g of Nickel? 1. Determine the molar mass of Ni mole g 2. Calculate the moles of Ni g Ni =.112 moles Ni moles Ni 58.69g Ni

11 Example #2: How many moles are in 92.2 g of FeO? 1. Determine the molar mass of FeO g/mole g/mole) g/mole 2. Calculate the moles of FeO g Fe moles FeO 71.85g FeO = 1.28 moles FeO The molar mass of an element or compound is used to convert moles of a substance into mass.

12 Example #3: How many grams are in 3.84 moles of NO 2? 1. Determine the molar mass of NO g/mole + 2(16.00) g/mole) 46.01g/mole 2. Calculate the moles of NO moles NO g NO moles NO 2 2 = 177 g NO 2

13 Percent Composition All substances are composed of elements. Those substances that contain several elements, do so with specific amounts of each component element. For example, consider K 2 CrO 4 and K 2 Cr 2 O 7 : The relative amounts are element expressed can be expressed as the Percent Composition.

14 The percent composition will be determined as a function of mass of each element in a compound. If a compound, AB, is made of elements A & B, then the massa % A x mass AB 100 Example #1: A compound is composed of g of sodium and g of chlorine. What is the percent composition? Answer: g g g

15 45.98 g % Na x g g % Cl x g OR % 60.66% % Cl = 100% % = 60.66%) Example #2: An 8.20 g piece of magnesium combines with 5.40 g of oxygen to form a compound. What is the percent composition of the compound? 8.20 g g g

16 8.20 g % Mg x g % 5.40 g % O x g OR % (% O = 100% % = 39.7%) If information is not given about the individual amounts for each element, the percent composition can be determined from the chemical formula. Example #1: What is the percent composition of C 3 H 8?

17 Answer: Step 1: Determine the molar mass of C 3 H 8. 8(1.01) g/mole + 3(12.01) g/mole) 44.11g/mole Step 2: Determine the molar masses for each type of element. For C: 3(12.01 g/mole) = g/mole For H: 8(1.01 g/mole) = 8.08 g/mole Step 3: Determine the percent composition for each element g % C x g % % H = 100% % = 18.32%

18 Example #2: What is the percent composition of HCN? Answer: Step 1: Determine the molar mass of HCN g/mole g/mole g/mole 27.03g/mole Step 2: Determine the percent compositions for each element. % C g x g % N g x g % 51.83% % H = 100% % %=3.74%

19 Example: What is the percent water in the hydrate cobalt (II) chloride dehydrate (CoC1 2 2H 2 O)? Answer: Step 1: Determine the molar mass of CoC1 2 2H 2 O. + 1(58.93 g/mole) 2(35.45 g/mole) 4(1.01 g/mole) 2(16.00 g/mole) g/mole Step 2: Determine the percent water. 4(1.01) 2(16.00) g / mole % H 2 O x g / mole 100 = % water

20 EMPIRICAL FORMULA The simplest ratio of the atoms in a molecule. Sometimes the molecular formula is the same as the empirical formula. Example: NaCl NaCl Molecular Formula Sometimes they are not: Empirical Formula C 3 H 9 CH 3 Molecular Empirical Formula Formula

21 Example #1: A compound has a composition g P 8.90 g H What is the empirical formula? Answer: 1) Convert the mass to moles mole P g P g P mole P mole H g H g H mole H 2) Select the atoms with the least number of moles. In this case it is the mole P

22 3) Divide each of the number of moles calculated in step #1 by the # of moles determined in step #2. For For P : H : Giving the ratio of P : H = 1 : 3 Empirical Formula PH 3

23 Example #2: A compound has a % composition 65.2% Sc 34.8% O What is the empirical formula? 1) Convert the percent to mass. 65.2% Sc 65.2 g Sc % O g O 100.0% g 2) Convert the mass to moles. 1 mole Sc 65.2 g Sc g Sc mole Sc mole O g O g O mole O 3) Select the atoms with the least number of moles. In this case it is the 1.45 mole Sc

24 4) Divide each of the number of moles calculated in step #2 by the # of moles determined in step #3. For For 1.45 Sc: O : Giving the ratio of Sc : O = 1 : 1.5 Empirical Formula Sc 1 O 1.5 Is this formula possible? Can you have 1.5 atoms??? NO! Solution: Multiply the subscripts in order to get whole numbers. Multiply by two. Empirical Formula Sc 2 O 3

25 Example #3: A compound is composed of 0.59 g H and 9.40 g O. It has been determined that the molar mass of the compound is 34.0 g/mole. 1. What is the empirical formula? 2. What is the molecular formula? Answer: 1. Empirical Formula 1 mole O 9.40 g O g O 0.59 For For 1 mole H g H g H 0.58 H : O : Giving the ratio of H : O = 1 : 1 mole mole H O

26 Empirical Formula HO 2. Molecular Formula The molar mass of the empirical formula: H O (1.01 g/mole) + (16.00 g/mole) = g/mole for HO If the empirical molar mass is doubled, then it will equal the molar mass of the molecular compound. Therefore, the molecular formula must be double the amount of atoms. Giving: H 2 O 2