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Dimensional Analysis

Dimensional Analysis. Conversion factors Units Cancel to solve. Counting Atoms. Mg burns in air (O 2 ) to produce white magnesium oxide, MgO. How can we figure out how much oxide is produced from a given mass of Mg?. Counting Atoms.

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Dimensional Analysis

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  1. Dimensional Analysis • Conversion factors • Units • Cancel to solve

  2. Counting Atoms Mg burns in air (O2) to produce white magnesium oxide, MgO. How can we figure out how much oxide is produced from a given mass of Mg?

  3. Counting Atoms Chemistry is a quantitative science—we need a “counting unit.” MOLE 1 mole is the amount of substance that contains as many particles (atoms, molecules) as there are in 12.0 g of 12C.

  4. Particles in a Mole Avogadro’s Number Amedeo Avogadro 1776-1856 6.02214199 x 1023 There is Avogadro’s number of particles in a mole of any substance.

  5. 1 mole 6.02 x 1023 representative particles 1 mole Si 6.02 x 1023 atoms Si moles = representative particles x moles = 2.80 x 1024 atoms Si x (atoms/molecules/formula units) (atoms/molecules/formula units) Converting particles to moles = 4.65 moles Si

  6. Rep part = moles x 6.02 x 1023 representative particles 1 mole Converting moles to particles 6.02 x 1023 molecule 1 mole Rep part = 1.14 mol SO3 x = 2.75 x 1024 molecules SO3

  7. The mass of moles • The atomic mass of an element expressed in grams is the mass of a mole of the element. • Molar Mass - the mass of a mole of an element. • Molar mass of a compound is the total mass of all the atoms making up the compound. • Molar mass of a compound is calculated by adding the masses of each element in the compound. • Take the subscript and multiply it by the molar mass of that element then add all the elemental masses together.

  8. Molar Mass 1 mol of 12C = 12.00 g of C = 6.022 x 1023 atoms of C 12.00 g of 12C is its MOLAR MASS Taking into account all of the isotopes of C, the molar mass of C is 12.011 g/mol

  9. One-mole Amounts

  10. PROBLEM: What amount of Mg is represented by 0.200 g? How many atoms? Mg has a molar mass of 24.3050 g/mol. How many atoms in this piece of Mg? = 4.95 x 1021 atoms Mg

  11. mass (grams) 1 mole mass = number of moles x 22.98977grams Na 1 mole Na mass = 3.00 mol Na x Converting moles to mass = 69.0 grams Na

  12. 1 mole mass (grams) moles = mass in grams x 1 mole Na2SO4 142.1 grams Na2SO4 moles = 10.0 g Na2SO4 x Converting mass to moles = 0.070 g Na2SO4

  13. Mole to volume ONLY GASES!!!!! • Avogadro’s hypothesis states equal volumes of gases at the same temperature and pressure contain equal numbers of particles. • STP – 0oC, 273 K / 1 atm, 101.3 kPa • Molar volume at STP • 22.4 L/mol

  14. 22.4 L 1 mol at STP volume = moles of gas x 22.4 L 1 mol at STP volume = 0.375moles of O2 x Converting mole to volume = 8.40 L O2

  15. 1 mol 22.4 L at STP moles of gas = volume x 1 mol 22.4 L at STP moles = 0.200 L H2 x Converting volume to moles = 8.93 x 10-3 L H2

  16. Grams = grams22.4 L x Moles L 1 mol Converting Molar Mass from Density • Density is mass per unit volume (g/L) • Molar mass grams per mol • Need to cancel moles and insert liters • Use molar volume – mol/liter

  17. molar mass = 1.964 grams22.4 L x (g/mol) 1L 1 mol Grams = grams22.4 L x Moles L 1 mol Converting Molar Mass from Density = 43.99 g/mol

  18. • • • • • • liters  The Molar Diagram molar volume 22.4 L/mol  Mol grams molar mass Periodic table Avogadro’s number 6.02 x 1023 particles/mol  atoms

  19. Percent Composition • Relative amounts (mass) of elements in a compound. • Percent by mass. • Mass of the element divided by the mass of the compound times 100. • Use the subscripts to determine the mass of each element and the mass of the compound if the question calls for a percent composition and only the formula is given.

  20. Percent Composition Consider NO2, Molar mass = ? What is the weight percent of N and of O? What are the weight percentages of N and O in NO?

  21. Determining Formulas In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICALor SIMPLESTformula. Remember formulas are mole to mole ratios. PROBLEM: A compound of B and H is 81.10% B. What is its empirical formula?

  22. A compound of B and H is 81.10% B. What is its empirical formula? • Because it contains only B and H, it must contain 18.90% H. • Assume a 100 gram sample. • In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. • Calculate the number of moles of each constituents.

  23. A compound of B and H is 81.10% B. What is its empirical formula? Calculate the number of moles of each element in 100.0 g of sample.

  24. A compound of B and H is 81.10% B. What is its empirical formula? Now, recognize that atoms combine in the ratio of small whole numbers. 1 atom B + 3 atoms H --> 1 molecule BH3 or 1 mol B atoms + 3 mol H atoms ---> 1 mol BH3 molecules Find the ratio of moles of elements in the compound.

  25. A compound of B and H is 81.10% B. What is its empirical formula? Take the ratio of moles of B and H. Always divide by the smaller number. But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B2H5

  26. B2H6 B2H6 is one example of this class of compounds. A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula? Is the molecular formula B2H5, B4H10, B6H15, B8H20, etc.?

  27. Molecular formulas • Molecular formulas are either the empirical formula or it is a whole number multiple of its empirical formula. • Determine the empirical formula first, to determine the molar mass of the compound. • Divide the experimental molar mass by the empirical molar mass. This give you the multiplier to convert the empirical formula to the molecular formula.

  28. A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula? We need to do an EXPERIMENT to find the MOLAR MASS. This experiment gives 53.3 g/mol Compare with the mass of B2H5 = 26.66 g/unit Find the ratio of these masses. Molecular formula = B4H10

  29. Determine the formula of a compound of Sn and I using the following data. • Reaction of Sn and I2 is done using excess Sn. • Mass of Sn in the beginning = 1.056 g • Mass of iodine (I2) used = 1.947 g • Mass of Sn remaining = 0.601 g • See p. 104

  30. Tin and Iodine Compound Find the mass of Sn that combined with 1.947 g I2. Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find moles of Sn used:

  31. Tin and Iodine Compound Now find the number of moles of I2 that combined with 3.83 x 10-3 mol Sn. Mass of I2 used was 1.947 g. How many mol of iodine atoms? = 1.534 x 10-2 mol I atoms

  32. Tin and Iodine Compound Now find the ratio of number of moles of moles of I and Sn that combined. Empirical formula is SnI4

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