Mole and Stoichiometry

Measuring Matter A mole is an amount of a substance Similar to: 12 in a dozen, 100 pennies in a dollar, etc. A mole of any substance represents 6.02 x 1023 particles of that substance 6.02 x 1023 = Avogadro’s Number 602000000000000000000000 6.02 x 1023 X atoms OR 1 mol X 1 mol X 6.02 x 1023 X atoms NOTE: the words atoms/molecules/particles are all the same

Problem 1: If you had 1 mole of donuts, how many donuts would you have? 6.02 x 1023  Problem 2: In 4.0 moles of glucose (C6H12O6), how many molecules/particles are there? 4.0 mol glucose x 6.02 x 1023 particles 1 mole glucose = 2.4 x 1024 particles

Molar Mass (formula mass) Must know how to determine molar mass (formula mass) Molar mass measures the mass of one mole of any element or compound The molar mass can be calculated using the atomic masses of each element (1 amu = 1 gram) from the periodic table 1 bromine atom = 79.90 amu Molar mass can be used to convert grams of a substance into moles or vice versa

Calculating the Molar Mass Molar Mass: add up all the amu values (make sure you look at subscripts!) What is the molar mass of 1 mole of H2O H: 2 x 1.01 = 2.02 O: 1 x 16.00 = 16.00 1 mole H2O = 18.02 g What is the molar mass of 1 mole of Cu(OH)2 Cu: 1 x 63.55 = 63.55 O: 2 x 16.00 = 32.00 1 mole Cu(OH)2 = 97.57 g

YOU TRY! Find the molar mass of the following: A) H3PO4 B) N2O5 (don’t forget to make sure neutral!) C) copper (II) nitrate D) potassium hydroxide

Calculations with mass/grams (amu) Mass of sample x conversion factor = number of moles in sample Calculate the mass of a sample that contains 23 moles of nitrogen (given: 23 moles N, unknown: N mass) 1 mol N= 14.01 grams 23 mole N x 14.01 grams = 322.23 grams N = 1 mol N 320g Calculate the number of oxygen gas moles (O2) present in a sample that has a mass of 290 grams. 1 mol O2 = 32.00 grams 290 grams x 1 mol O2 = 9.0625 moles O2 = 32 grams 9.1 moles

More molar mass problems YOU TRY! A) How many moles are in 250 grams of water? B) How many grams are in 3.21 moles of C12H22O11?

Calculations with moles and particles particles x conversion factor = number of moles Chromium is a metal that is added to steel to improve its resistance to corrosion. Calculate the number of moles in a sample of chromium containing 5.00 x 1020 atoms. 5.00 x 1020 atoms Cr x 1 mol Cr = 8.31 x 10-4 mol Cr 6.02x1023atoms Cr

More molar mass problems YOU TRY! A) How many molecules are in 2.32 moles of HNO3? B) How many moles are in 3.60 x 1024 molecules of Li2O?

Calculations with grams to particles grams x conversion factor = moles; moles x conversion factor = particles Calculate the number of salt molecules present in a sample of 63.2 grams of salt 63.2 g NaCl x 1 mol NaCl x 6.02 x 1023 molecules 58.44 g NaCl 1 mol NaCl = 6.5103 x 1023 = 6.51 x 1023 molecules

More molar mass problems YOU TRY! (two steps!) A) Calculate the number of iron (Fe) atoms present in a sample that has a mass of 4021.2 grams. B) If a sample of sodium chloride (NaCl) contains 8.2 x 1026 particles, how many grams of NaCl are in the sample?

Percent by Mass Composition When given a sample of a compound, you can determine how much of each element is present Percent composition is the percent by mass of each element in a compound The following formula is used to calculate percent composition: % Mass of Element  =  (grams of element/grams of compound) x 100%

Percent Composition Example: Calculate the percentage of carbon in ethane, C2H6 grams of carbon = 2 x 12 = 24 grams grams of compound = 24 + (6 x 1) = 30 grams (24 /30)  x 100  =  80% Example: Find the percentage composition of a compound that contains 2.7369 g of chlorine, 0.4116 g of oxygen, and 0.7971 g of phosphorus in a 3.9460 g sample of the compound Cl (2.7369 g/3.9460 g) x 100 = 69.4% O (0.4116 g/3.9460 g) x 100 = 10.4% P (0.7971 g/3.9460 g) x 100 = 20.2%

Percent Composition Example: A sample of an unknown compound with mass of 2.876 g has the following composition: 66.07% carbon, 6.71% hydrogen, 4.06% nitrogen, 23.16% oxygen. What is the mass of each element? Equation: (% Mass x grams compound) *you can use other methods 100% C (66.07x2.876 g)/100 = 1.90g H (6.71x2.876 g)/100 = .193g N (4.06x2.876 g)/100 = .117g O (23.16x2.876 g)/100 = .66g

YOU TRY! A) Calculate the percent by mass composition of dimethylether, CH3OCH3 B) What is the percent composition in each element of Iron (III) oxide?

Empirical Formula The empirical formula gives the simplest whole number ratio of the elements in a compound Example: empirical formula for hydrogen peroxide, H2O2, is HO. The ratio is 1:1 To determine: 1) convert mass of each element given into moles using molar mass (from periodic table) 2) divide each mole value by the smallest # of moles 3) round the mole ratio of the elements to the nearest whole # (NOTE: if something rounds to .5 then you have to multiply everything by 2) NOTE: Sometimes instead of mass it gives you a percent. Just call it grams.

Empirical Formula Example: A compound was found to contain 7.30 g Na, 5.08 g S, and 7.62 g O. Determine the empirical formula. 7.30 g Na x 1 mol Na = .317 mol Na 5.08 g S x 1 mol S = .158 mol S 23.0 g Na 32.1 g S 7.62 g O x 1 mol O = .476 mol O 16.0 g O SMALLEST # MOLES = .158 Na .317/.158 = 2.01 S .158/.158 = 1.0 O .476/.158 = 3.0 Answer: Na2SO3 (Also accept: Na2O3S)

Empirical Formula Example: Determine the empirical formula of a compound containing 20.23% aluminum and 79.77% chlorine Al 20.23%: 20.23 g Al x 1 mol Al = .750 26.98 g Al Cl 79.77%: 79.77 g Cl x 1 mol Cl = 2.25 35.45 g Cl SMALLEST # MOLES = .750 Al .750/.750 = 1.0 Cl 2.25/.750 = 3.0 Answer: AlCl3

Empirical Formula YOU TRY! A hydrocarbon (a compound containing only hydrogen & carbon) is found to be 7.690% H and 92.31% C by mass. Calculate its empirical formula. A 170.00 g sample of an unidentified compound contains 29.84 g sodium, 67.49 g chromium, and 72.67 g oxygen. What is the compound’s empirical formula?

Molecular Formula The actual number of atoms in a compound A whole-number multiple of the empirical formula CH2O; C6H12O6; C12H24O12 First find the empirical formula, then divide the molar mass of the entire compound by the mass of the empirical formula to determine the multiple

Molecular Formula Example: Ribose has a molar mass of 150 g/mol and a chemical composition of 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen. What is the molecular formula for ribose? C: 40.0 gC x 1 mol C = 3.33 mol C H: 6.67 gH x 1 mol H = 6.60 mol H 12.0 g C 1.01 g H O: 53.3 gO x 1 mol O = 3.33 mol O 16.0 g O Empirical Formula= C1H2O1 empirical formula mass   30.0 g/mol (12+1+1+16) molar mass = 150 g/mol molecular formula = 150 / 30 = 5 x empirical C5H10O5

Molecular Formula YOU TRY! A) A hydrocarbon (a compound containing only hydrogen & carbon) is found to be 7.690% H and 92.31% C by mass. The molar mass is 78.12 g/mol. Calculate its molecular formula (use the previous empirical formula of CH).

Balancing Chemical Equations Balancing equations is done to satisfy the Law of Conservation of Mass to make the # of atoms the same on both sides of the equation Add coefficients but leave the subscripts alone Step by Step Instructions (until you become a professional): Draw a vertical line down from below the arrow in the equation. On EACH side of the line, list all of the elements that appear. Do not repeat elements on the same side. For EACH side of the line, count the number of atoms for each element. Add coefficients in front of formulas so that the number of atoms on each side matches. Once all element counts match, make sure all coefficients are reduced to their lowest whole-number ratio.

Balancing equations This process in rather unscientific and is essentially a process of trial and error TIP #1: If an element appears in only one compound on each side of the equation, try balancing that first. TIP #2: If one of the reactants or products appears as the free element, try balancing that last. TIP #3: Remember that you can only change the coefficient. NOTE: coefficients should be whole numbers

YOU TRY! Balance the following: A) Fe + Br2  FeBr3 B) C4H8 + O2  CO2 + H2O C) Pb(NO3)2  PbO + NO2 + O2

Mole Ratios Balanced equation: C3H8 +5O2  3CO2 + 4H2O is really (1 mol C3H8 +5 mol O2 yields 3 mol CO2 + 4 mol H2O) What number of moles of CO2 will be produced by the decomposition of 3.5 mol of C3H8? Unknown = moles of CO2 Given = 3.5 moles C3H8 Also, 1 mol C3H8 = 3 mol CO2 3.5 mol C3H8 x 3 mol CO2 = 10.5 mol CO2 = 11 mol 1 mol C3H8

Another example Solid iron reacts with oxygen to produce iron(III) oxide. If you wanted to produce 2.5 moles of iron(III) oxide, what mass of oxygen must react? 4Fe + 3O2  2Fe2O3 Unknown = Grams of O2 Given = 2.5 Moles Fe2O3 2.5 moles Fe2O3 3 mol O2 32 grams O2 2 mol Fe2O3 1 mol O2 Answer: 120 grams O2 (2.5 x 3 x 32) / 2

Mole Ratios YOU TRY! CH4 + 2O2  CO2 + 2H2O Calculate the grams of O2 required to produce 2.23 mol of carbon dioxide. Calculate the mass of water produced when 34.0 g of CH4 is burned.