# Mole and Molar Mass - DOC

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```					Chemistry

Guided Learning Module

Spring, 2006

Mole, Molar Mass, and Molarity
Model: Definition of a Mole and the Molar Mass Key Questions 1. In your own words, what are the definitions of the terms: mole and molar mass? Mole – A mole is a counting unit, i.e. a name for a uniform number of objects. Just as a dozen equals twelve items, and ream equals 500 items, and a googol equals 1.0 x 10100 items, a mole equals 6.02 x 1023 items. Molar mass – the mass in grams of a mole of items with a defined mass. 2. What determines the value of Avogadro’s number? Avogadro’s number is equal to the number of carbon-12 atoms that would be found in a sample with a mass equal to exactly 12 grams. 3. How many apples are in a dozen apples, and how many apples are in a mole of apples? There are 12 apples in a dozen apples. There are 6.02 x 1023 apples in a mole of apples. 4. How can you determine the mass of one carbon-12 atom from the information in the model? You can determine the mass of one carbon-12 atom by dividing 12.0 grams/mole (the weight of one mole of carbon-12 atoms) by Avogadro’s number (the number of items, in this case atoms, in one mole). 12.0 grams per mole / 6.02 x 1023 atoms per mole = 1.99 x 10-23 grams/atom 5. What are the unit conversion factors for changing; a. moles of an element to number of atoms. The conversion factor is Avogadro’s number 6.02 x 1023 atoms/mole. You would multiply the number of moles by Avogadro’s number to get the number of atoms.
b.

grams of an element to moles.

The conversion factor is 1/molar mass (which has units of grams per mole). You would multiply the number of grams by 1/molar mass to get the number of moles. c. grams of an element to number of atoms. You would perform step b, and then step a. Multiply grams by 1/molar mass and by Avogadro’s number. e3ef44b2-8dd2-417c-859a-d969942ee8e5.doc 10/31/2009 9:59:00 AM 1

Chemistry

Guided Learning Module

Spring, 2006

6. Why are abundance-weighted average molar masses listed in tables of the elements? Abundance-weighted average molar masses are listed in tables of the elements because they represent the mass of a naturally occurring sample of an element. 7. When might it be useful to know how many atoms or molecules are present in a sample of material? It would be useful to know how many atoms or molecules are present in a sample if you were trying to figure out exactly how much product you would be able to produce in a chemical reaction or determine the stoichiometric ratio of elements in a compound. Exercises 1. If the average mass of an apple is 0.15 kg, calculate a. the mass of a dozen apples. 0.15 kg/apple x 12 apples/dozen apples = 1.8 kg/dozen apples b. the mass of a mole of apples. 0.15 kg/apple x 6.02 x 1023 apples/mole apples = 0.90 x 1023 kg/mole apples = 9.0 x 1022 kg/mole apples 2. Calculate the number of atoms in a 6.00 g sample of carbon-12. 6.00 g carbon-12 x 1 mole /12.0 grams carbon-12 x 6.02 x 1023 atoms/mole = 3.01 x 1023 atoms 3. Calculate the mass in grams of one carbon-12 atom. 12.0 grams / mole carbon-12 x 1 mole / 6.02 x 1023 atoms = 1.99 x 1023 grams / carbon-12 atom 4. Calculate the number of moles in 75 g of iron. 75 grams x 1 mole Fe/ 55.85 grams = 1.3 moles Fe 5. Calculate the number of atoms in 0.25 moles of uranium. 0.25 moles U x 6.02 x 1023 atoms/mole = 1.5 x 1023 atoms U 6. Calculate the number of moles corresponding to 12.04 x 1023 atoms of uranium. 12.04 x 1023 atoms U x 1 mole/6.02 x 1023 atoms = 2.00 moles U 7. Calculate the mass of 12.04 x 1023 atoms of uranium. 12.04 x 1023 atoms U x 1 mole/6.02 x 1023 atoms x 238 grams/mole U = 476 grams

e3ef44b2-8dd2-417c-859a-d969942ee8e5.doc 10/31/2009 9:59:00 AM

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Chemistry Got It!

Guided Learning Module

Spring, 2006

1. If you have 1 g samples of several different chemical compounds, which sample will contain the fewest molecules? The sample of the compound that has the highest molar mass will have the fewest molecules per 1 gram sample. Problems. 1. A mass of 30.0 g of oxygen reacts with 6.02214 x 1023 atoms of carbon. a. What is the molar ratio of carbon to oxygen in the product? Moles of oxygen = 30.0 g oxygen x 1 mole oxygen/15.999 g oxygen = 1.88 moles oxygen Moles of carbon = 6.02214 x 1023 atoms carbon x 1 mole / 6.02214 x 1023 atoms = 1.00 mole carbon Molar ratio carbon/oxygen = 1.00 moles carbon/1.88 moles oxygen = 0.532 moles carbon/moles oxygen b. Given that mass is conserved in a chemical reaction, what is the mass of the product produced? Mass product = mass reactants Mass product = mass C + mass O = 1.00 mole x 12.01 g/mol + 30.0 g oxygen = 42.0 grams c. Is the product carbon monoxide, CO, or carbon dioxide, CO2? It is probably a mixture of both. Model: Beads in a Jar - Average Mass of a Mixture of Objects Key Questions 8. How can you estimate the average mass of a bead in the jar by examining the data in the model in order to identify whether option a, b, or c below is the best answer? By inspection, the mass of the heavier 50% of the beads is going to be < 2.75 g, because more than half the beads are 2.5 g. The average of 2.00 g and <2.75 g is going to be <2.375 g. which is choice b. 9. How can you calculate the average mass of a bead in the jar from the data in the model? The average mass can be calculated as a weighted average. Exercises 8. Calculate the average mass of a bead in the jar. ((50 x 2.0 g) + (30 x 2.5 g) + (20 x 3.0 g))/100 = 2.35 grams

e3ef44b2-8dd2-417c-859a-d969942ee8e5.doc 10/31/2009 9:59:00 AM

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Chemistry Problems

Guided Learning Module

Spring, 2006

2. The molar mass of 35Cl is 34.971 g. and the molar mass of 37Cl is 36.970 g. In a natural sample, 75.77% of the atoms are 35Cl, and 24.23% are 37Cl. Show how to calculate the average molar mass of a natural sample of chlorine. (75.77% x 34.971 g/mole + 24.23% x 36.970 g/mole )/100% = 35.455 g/mole Model: Recipes for Preparing Common Molar Solutions: Key Questions 10. How many moles of NaCl did 1 Liter of a 1.00 M NaCl solution contain? Molar mass of NaCl = 23.0 g/mol Na + 35.5 g/mol Cl = 58.5 g/mole NaCl 58.5 g NaCl added x 1 mole NaCl/58.5 g NaCl = 1.00 mole contained in l liter of solution. 11. How many moles of MgO were added to give 500 ml of a 1.00 M solution? Molar mass of MgO = 24.3 g/mol Mg + 16.0 g/mol O = 40.3 g/mole MgO 20.2 g MgO added x 1 mole MgO/40.3 g MgO = 0.501 mol MgO

12. How many moles would you add to make 500 ml of a 5 M solution of MgO? 5 M MgO = 5 mol MgO/1 Liter x 500 ml x 1 liter/1000 ml = 2.5 mol MgO 13. How many grams would you need to make the solution in the previous question? 40.3 g MgO/mole MgO x 2.5 mole MgO = 1.0 x 102 grams 14. How many moles of NaOH would you need to make 1 L of 2 mM NaOH? 2 mM NaOH = 2 mmoles NaOH/1 liter solution x 1 liter = 2 mmoles NaOH 15. How many grams of NaOH would the answer to question 14 be? 2 mmoles NaOH = 2 x 10-3 moles NaOH Molar mass of NaOH = 23.0 g Na/mole Na + 16.0 g O/mole O + 1.00 g H/mole H = 40.0 g NaOH/mole NaOH 2 x 10-3 moles NaOH x 40.0 g NaOH/mole NaOH = 80 x 10-3 g = 80 mg

e3ef44b2-8dd2-417c-859a-d969942ee8e5.doc 10/31/2009 9:59:00 AM

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Chemistry Problems

Guided Learning Module

Spring, 2006

3. You need to make at least 2.0 liters of a 0.5 M CaCl2 solution. Describe what you would do. First, calculate the number of grams of CaCl2 required: 0.5 M CaCl2 = 0.5 mol CaCl2 / 1 liter molar mass CaCl2 = 40.1 + 2(35.5) = 111.1 grams/mol 0.5 mol CaCl2 / 1 liter x 2 liters = 1.0 mol CaCl2 1.0 mol CaCl2 x 111.1 g/mol = 111.1 g CaCl2 Mass out 111.1 g of CaCl2 and dissolve in about 1.8 liters of H2O in a 2 liter graduated cylinder. Add H2O to bring total volume to 2.0 liters. 4. What volume would you have to dissolve 100 g of NaOH into to obtain a 2 uM solution? 100 g NaOH x 1 mol/40.0 g = 2.5 mol NaOH 2.5 mol NaOH x 1 liter/2 x 10-6 mol = 1.25 x 106 liters = ~1 x 106 liters
5.

How many moles of NaCl would you have if you transferred 0.1 ml of a 1 M NaCl solution to a test tube?

1 M NaCl = 1 mol NaCl/liter x 1 liter/1000 ml x 0.1 ml = 1 x 10-4 mol NaCl Got It! 2. A titration is a laboratory procedure in which a measured volume of a solution whose concentration is known is added to a known volume of a second solution until an end-point is reached. For example, if a 25.0 ml of a 1 M HCl solution were required to neutralize 50.0 ml sample of a NaOH solution, the concentration of the original NaOH solution can be determined. What is the concentration of the NaOH solution? First calculate the number of moles of HCl added to neutralize: 1 M HCl = 1 mol HCl/1000 ml x 25.0 ml = 2.50 x 10-2 mol HCl At neutrality, the number of moles of H+ = the number of moles of OH- present, therefore: 2.50 x 10-2 mol H+ = 2.50 x 10-2 mol OHSince the 2.50 x 10-2 mol OH- was contained in 50.0 ml of solution, the molarity of the NaOH solution was: 2.50 x 10-2 mol OH- /50.0 ml x 1000 ml/liter = 0.5 M NaOH

e3ef44b2-8dd2-417c-859a-d969942ee8e5.doc 10/31/2009 9:59:00 AM

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