Dimensional Analysis Conversion factors Units Cancel to solve

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?

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.

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.

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

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

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.

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

One-mole Amounts

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

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

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

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

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

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

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 Grams = grams 22.4 L x Moles L 1 mol

Converting Molar Mass from Density Grams = grams 22.4 L x Moles L 1 mol = 1.964 grams 22.4 L x 1L 1 mol molar mass (g/mol) = 43.99 g/mol

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

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.

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?

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

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.

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.

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.

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

A compound of B and H is 81. 10% B. Its empirical formula is B2H5 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.? B2H6 is one example of this class of compounds. B2H6

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.

A compound of B and H is 81. 10% B. Its empirical formula is B2H5 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

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

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:

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

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