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Molar Mass and Formulas

Molar Mass and Formulas. Chapter 3 (sections 3.1-3.5). What is a mole?. A mole is used to represent the quantity 6.02 x 10 23 . It can be used to represent that amount of ANYTHING: -atoms -molecules -ions -compounds -particles -items (ex: students, homework assignments).

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Molar Mass and Formulas

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  1. Molar Mass and Formulas Chapter 3 (sections 3.1-3.5)

  2. What is a mole? • A mole is used to represent the quantity 6.02 x 1023. It can be used to represent that amount of ANYTHING: -atoms -molecules -ions -compounds -particles -items (ex: students, homework assignments)

  3. How is a mole defined? • One mole is defined as the number equal to the number of carbon atoms in exactly 12 grams of pure 12C. This amount has been determined as precisely 6.02214 x1023 atoms. • Moles are convenient to use in a chemistry lab because they represent quantities that are typically easy to see, handle and measure. For example, one mole of water molecules weighs about 18 grams, or has a volume of about 18 mL.

  4. Avogadro’s number • This number is also called Avogadro’s number (NA). For a good time call 602-1023

  5. Calculating Molar Mass • Molar mass is the mass of 1 mole of a pure substance. • Molar mass may also be referred to as: -gram formula mass (ionic compounds) -formula mass (ionic compounds) -molecular mass (non metals)

  6. Calculating Molar Mass • The molar mass of a formula can be determined by adding the molar masses of each of the elements present. EX: H2O Elements Moles of each Atomic mass H 2 mol x 1.01 g/mol = 2.02 g O 1 mol x 16.00 g/mol =16.00 g Molar mass of H2O  18.02 g/mol

  7. Practice Problem Calculate molar mass of the following: Al2S3 Al: 2 mol x 26.98 g/mol = 53.96 g S: 3 mol x 32.07 g/mol = 96.21 g 150.17 g Al2S3/mol

  8. atoms/molecules Mole Conversion Map 6.022x1023 molec/mol equation coeffients grams moles (of something else) formula mass MOLES 22.4 L/mol volume of a gas at STP ***remember when using formula mass, it is always for only ONE mole

  9. Practice Conversions 1. What is the mass in grams of 2.5 mol of O2? 2.50 mol O2 x 32.00 g O2= 1 mol O2 2. Determine the number of moles in 5.00 g of H2O: 5.00 g H2O x 1 mol H2O = 18.02 g H2O 80.0 g of O2 0.277 mol of H2O

  10. Multiple Conversions 3. Determine the number of molecules in a 5.45 g sample of CaCl2. 5.45 g CaCl2 x 1 mol CaCl2 x 6.02 x 1023 molec. CaCl2= 110.98 g CaCl2 1 mol CaCl2 2.96 x 1022 molecules of CaCl2

  11. Percent Composition • Percent composition of a compound is a statement of the relative mass each element contributes to the mass of the compound as a whole. • Chemists often compare the percent compositions of unknown compounds to those of known compounds to identify the unknown by its ratio of elements, or chemical formula.

  12. Calculating % Composition Example: Find the % composition of the elements in NaCl. Step 1: calculate the molar mass of the substance. Molar mass of NaCl is 58.44 g/mol Step 2: divide the molar mass of each of the elements by the molar mass of the substance. Step 3: multiply by 100% to get percent. Na = 22.99g/mol x 100 = 58.44g/mol Cl = 35.45g/mol x 100 = 58.44g/mol 39.3% Na 60.7% Cl

  13. Empirical Formulas • Remember that the elements in a compound combine in whole number ratios such as 1:1, 1:2, 2:3, and so on. • If elements combine in these whole number ratios, we can predict the same applies for moles of each atom. • We can use this principle to find the empirical (lowest whole number) formula for compounds based on the relative masses of each of the elements in the compound.

  14. Calculating Empirical Formulas Example: What is the empirical formula for a compound if a 2.50 gram sample contains 0.900 g of calcium and 1.60 g of chlorine? Step 1: convert grams of each of the elements into moles. 0.900 g Ca x 1 mol Ca = 0.0224 mol Ca 40.08 g Ca 1.60 g Cl x 1 mol Cl = 0.0451 mol Cl 35.45 g Cl Step 2: obtain the simplest ratio by dividing the moles by the smallest number of moles. 0.0224 mol Ca/0.0224 mol Ca = 1.00 ratio 0.0451 mol Cl/0.0224 mol Ca = 2.01 ratio Step 3: round or multiply to express the ratio using whole numbers. CaCl2

  15. Practice Problem What is the empirical formula of a compound that is 66.0% Ca and 34.0% P? *NOTE: When given only the percentages of each element, assume that there is 100 g of the compound. *NOTE 2: Sometimes dividing by the smallest number of moles will not lead to a whole number. You must multiply the ratio by a whole number so that the ratio is within 10% of a whole number.

  16. ANSWER: Ca: 66.0 g Ca x 1 mole Ca = 1.65 mol Ca 40.08 g Ca P: 34.0 g P x 1 mole P = 1.10 mol P 30.97 g P Ca: 1.65 mol Ca/1.10 mol P = 1.50 P: 1.10 mol P/1.10 mol P = 1.00 Ca: 1.50 x 2 = 3.00 P: 1.00 x 2 = 2.00 Ca3P2

  17. Furnace sample CO2, H2O, O2 and other gases O2 in O2 and other gases such as Mg(ClO4)2 such as NaOH Combustion analysis A typical method for determining the empirical formula of a compound is by decomposing it by combustion. When an organic compound is burned in the presence of oxygen, the CO2, H2O and N2 produced can be collected and weighed. The results can be used to determine the mass percent of each element.

  18. Practice Problem • When 0.1156 g of an unknown substance composed of C, H and N is burned, 0.1638g of CO2 and 0.1676g of H2O are collected. The molar mass of the substance is about 62.5 g/mol. Assume all of the C and H came from the unknown. Determine the empirical and molecular formulas.

  19. ANSWER: • C:0.1638 g CO2 x I mol CO2 X1 mol C x 12.01 g C = 0.04470 g C 44.01 g CO2 1 mol CO2 1 mol C • H: 0.1676 g H2O x 1 mol H2O x 2 mol H x 1.0008 g H = 0.01875 g H 18.02 g H2O 1 mol H2O 1 mol H • %C:0.04470 g C x 100% = 38.67% C 0.1156 g cmpd • %H: 0.01875 g H x 100% = 16.22% H 0.1156 g cmpd • %N: 100.00% - (38.67%C + 16.22%H) = 45.11% N

  20. ANSWER CONT… C: 38.67 g C x 1 mole C = 3.220 mol C 12.01 g C H: 16.22 g H x 1 mole H = 16.09 mol H 1.008 g H N: 45.11 g N x 1 mole N = 3.220 mol N 14.01 g N C: 3.220 mol C/3.220 mol C = 1.00 H: 16.09 mol H/3.220 mol C = 4.997 N: 3.220 mol N/3.220 mol C = 1.00 CH5N

  21. ANSWER CONT… • If the empirical formula is CH5N, then the molecular formula is a whole number multiple of this formula: • Molar mass CH5N = 31.07 g/mol • Ratio is 62.5 g/mol = 2.01 31.07 g/mol • Molecular formula = C2H10N2

  22. THE END

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