Molar Math I. AMU and Avogadro's Number II. Molar Mass A. GAM B.
GMM C. GFM III. Molar Volume of a gas IV. Calculation Circle V. %
Composition VI. Formulas A. Empirical B. Molecular C. Molecular /
Empirical Ratios VII. Empirical Formula from % Composition VIII.
Density of a gas
Molar Math I. AMU A. Definition - the standardized mass used to
mass atoms Equal to 1/12 the mass of C12 isotope B. Determination
1. Chemists use a mass spectrometer to determine the relative
masses and abundance of atoms (isotopes). a. The masses of atoms
are in a ratio to each other: C - 12 amu , He - 4 amu, H - 1
amu
b. It follows then that 12g C, 4g He and 1g H all have the same
number of particles. What is the number of particles associated
with these gram masses? c. Avogadro's number = 1 Mole = 6.02x1023
(1) How big is this number? 88.5 trillion times the population of
the Earth 1 mole of pennies would pay all the expenses of the US
for the next billion years. (2) Why is the mole so large?
2. Practice: a. 1/4 mole = b. 1/2 mole = c. 1 mole = 6.02x1023
d. 2 moles = e. 4 moles =
3. FORMULA
II. Molar Mass - the mass in grams of 1 mole of ..... A. Gram
Atomic Mass (gam)- the mass in grams of 1 mole of atoms - (I count
any decimals between .2 and .8 in atomic mass, everything else is
rounded to nearest whole #) 1. gam H = 1 g/mol 2. gam C = 12 g/mol
3. gam O = 16 g/mol 4. gam N = 14 g/mol 5. gam Na = 23 g/mol 6. gam
Cl = 35.5 g/mol B. Gram Molecular Mass (gmm)- the mass in grams of
1 mole of molecules - (simply add up all atomic masses in the
molecule) 1. gmm Hydrogen , H2 = 2g/mol 2. gmm O2 = 32 g/mol 3. gmm
Cl2 = 71 g/mol 4. gmm = He = 4 g/mol 5. gmm H2O = 18 g/mol 6. gmm
NH3 = 17 g/mol 7. gmm CO2 = 44 g/mol 8. gmm C6H12O6 = 180 g/mol C.
Gram Formula Mass (gfm)- the mass in grams of 1 mole of formula
units(ionic) 1. NaCl = 58.5 g/mol 2. CaCO3 = 100 g/mol 3. NH4NO3 =
80 g/mol D. Practice: Which has a greater number of particles?
58.5 g NaCl, 90 g C6H12O6, 4 g He, 48 g H2O
III. Molar Volume of a Gas A. One mole of any gas at STP
occupies 22.4 L. 1. why? Gases are composed of molecules that are
extremely far apart from each other (in relation to their size)
therefore the volume of the molecules is negligible compared to the
volume of the overall gas.
B. Practice: 1. 1 mol Cl2(g) = _____ L 2. .5 mol H2O(g) = _____
L 3. 3 moles NH3(g) = _____ L 4. 33.6 L He(g) = ______ mol 5. 2.2L
O2(g) = _____ mol IV. Mole Calculation Circle
A. multiply going out, divide going in!
V. % Composition The percent composition by mass of any
constituents is calculated using the following formula:
% Comp = (mass part / mass whole) x 100A. Examples: 1. % O in
H2O = (16g/mol / 18 g/mol) x 100 = 88.9% O % H in H2O = (2g/mol /
18 g/mol) x 100 = 11.19% H 2. % Cu in CuSO4 = (63.5g/mol /
159.5g/mol) x 100= 39.8% Cu % S in CuSO4 = (32g/mol / 159.5g/mol) x
100= 20.1% S % O in CuSO4 = (4(16g/mol) / 159.5g/mol) x 100= 40.1%
O B. Practice: 1. % H2O in MgCl2 . 3H20 2. % C, %H, %O in C2H5OH 3.
%O in iron III dichromate tetrahydrate
VI. Formula Varieties A. Empirical Formula: A formula written in
lowest terms (lowest whole number ratio) Examples: 1. H2O 2. NH3 3.
CO2 4. NaCl 5. C6H12O6 - CH2O C3H15O9 - CH5O3 H2O2 - HO
B. Molecular Formula: The formula of a substance as it exists in
nature. (contains all altoms in the molecule/cmpd) 1. C6H12O6 2.
C6H6 3. H2O2 4. H2O 5. NH3 C. Molecular / Empirical Ratios The
molecular formula is always some multiple of the empirical
formula.
Molecular Formula
= (Empirical Formula)x
1. Steps to take: a. Determine molar mass of the empirical
formula. b. Divide molecular mass / empirical mass to determine
ratio! c. Then, multiply empirical formula by ratio factor to
derive molecular formula. 2. Example: a. The empirical formula is
P2O5. Experiments have determined the formula mass to be
283.889g/mol. What is the molecular formula? What is it's name?
Step 1 - molar mass of P2O5= 137g/mol Step 2 - Determine Ratio:
283.889 g/mol / 137 g/mol = 2.07 or molecular formula = 2 x
empirical formulatetraphosphorus decoxide
Step 3 - Multiply through: 2(P2O5) = P4O10 3. Practice:
a. The empirical formula is CH, the molecuar mass is 78.110
g/mol. Determine the molecular formula. b. A sample contains 0.44 g
H and 6.92 g O. If its molecular mass = 34.00 g/mol what is it's
molecular formula?
VII. Empirical Formulas from % Composition A. Steps to Solve: 1.
turn % to grams (assume 100 gram sample) 2. convert grams to moles
3. determine ratio of atoms by dividing large # by small # 4. if
ratio is not whole # multiply by 2 B. Examples: 1. Find the
empirical formula of a compound that is 48.38% carbon, 8.12%
hydrogen, and 53.5% oxygen by mass. a. First turn % to grams b.
Second, turn grams to moles: (48.38 g C) (1 mol/ 12.10 g C) = 4.028
mol C (8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H (53.38 g O) (1
mol/ 16.00 g O) = 3.336 mol O c. Third, find the ratio or the moles
of each element by dividing the number of moles of each by the
smallest number of moles. (3.336 mol O/ 3.336) = 1 mol O (4.028
mol C/ 3.336) = 1.2 mol C (8.056 mol H/ 3.336) = 2.4 mol H d.
Lastly, multiply until you reach lowest whole numbers. (1 mol O)
(5) = 5 mol O (1.2 mol C) (5) = 6 mol C (2.4 mol H) (5) = 12 mol
H
C6H12O5
2. A compound has the following mass% composition. Determine the
empirical formula if Hg:91.3% and F:8.65%. a. First turn % to grams
b. Second, turn grams to moles: Hg: 91.3/200.59=0.4552 mol F:
8.65/18.9984=0.4553 mol c. Third, find the ratio or the moles of
each element by dividing the number of moles of each by the
smallest number of moles. Since they are both the same it is
clearly a 1:1 ratio, therefore the empirical formula is:
HgF
C. Practice problems: click here
VIII. Density of a gas By knowing the molar mass and molar
volume of a gas (at STP) it is possible to determine the density of
any gas,
For example: What is the density of Carbon Monoxide at STP? A.
Solution: The molar mass of CO is 12+16=28 g/mol.
Therefore D=M / V
D= 28g / 22.4L = 1.25 g/L
