The formula and name denote elements and relative composition in

CHEM 110: Peer Review, Ch 2-3 WS
Leah Swanzy - [email protected]
The formula and name denote elements and relative composition in the compound. A balanced equation conserves atoms
and moles of each element.
1 mole is Avogadro’s Number of atoms or molecules.
Avogadro’s # = 6.023 x 1023
Therefore we can say that 1 mole = 6.023 x 1023 atoms.
So 1 mol of O2 = _______________ molecules of O2 = _______________ atoms of oxygen
Molar mass of a compound is the mass in grams of 1 mole of the substance.
The molar mass of O2 = ________ g
Calculating Molar Mass:
Molar mass = sum of each element’s atomic weight multiplied by the coefficient.
Ex: MgCl2
1. Mg = ______ g/mol
Cl = 35.5 x 2 = ______ g/mol
2. Molar mass = the sum = 24.3 + 71 = ________ g/mol
Elemental Percent Composition is the portion of mass coming from each element in the compound. Remember that the
percent compositions must add to equal 100%.
Step 1: Find the number of atoms for each element
Step 2: Find the total mass of each element present = (atomic weight) x (# of atoms of that element)
Ex: Determine % Comp of Mg in MgCl2
1. MgCl2 = 95.3 g/mol (from above)
2. Mg = 1 atom
Cl = 2 atoms
3. Mass Mg = 1 x 24.3 = 24.3
Mass Cl = 2 x 35.5 = 71
4. Find % composition:
a. Mg: (
b. Cl: (
) = (100)  x = _____ % Mg
) = (100)  x = _____% Cl
5. Check your work: _______ + ______ = 100.00% 
Calculating # of molecules:
Ex: Determine the number of moles and the number of molecules in 5.0g of water vapor.
1. 5g H2O = ______g/mol (2+16=18)
2. # of Mols:
3. # of molecules:
CHEM 110: Peer Review, Ch 2-3 WS
Leah Swanzy - [email protected]
Empirical Formulas: chemical formula that indicates the relative proportions of the elements in a molecule rather than
the actual number of atoms of the elements. It is a ratio.
Ex: Determine the empirical formula for a compound containing 75% C and 25% H.
1. Assume 100g (make it easy for yourself)
a. This means there are _____g C and _____g H
2. Find the # of moles of each element
a. 75 g C
b. 25 g H
3. C 6.24 H 24.8  We want to reduce this down to a simple ratio. Divide by the lowest #:
The Empirical Formula is:
Now let’s do the reverse: find the molecular formula from the empirical formula. Remember that the molecular
coefficients are found by multiplying the entire empirical formula by a factor.
Ex: Determine the molecular formula
for a compound with molar mass 28 and empirical formula CH2.
1. Find the molar mass of the empirical formula:
2. Now we want to know what factor to multiply the empirical formula by. The goal is to have the molecular molar
mass = 28g (from the problem).
a. Molecular/Empirical = this factor
b. The factor is ___, so we multiply the entire empirical formula (all coefficients) by ___:
3. Answer:
Balancing Equations:
The goal is to have the same number of atoms per element on either side of the equation:
C + O2  CO2
On the left side, we have 1C and 2O. On the right, we also have 1C and 2O. This is balanced.
H2O  H2 + O2 This is not balanced. Who is incorrect?
1. Identify each element in the reaction.
2. Tally up how many of each element you have on either side of the equation. Remember that the coefficient
applies to each atom in a molecule or polyatomic ion. (2 H2O = __H and __O)
Example: __CH4 + __O2  ___CO2 + ___ H2O
Zinc has an atomic number of thirty. The average atomic mass is 65.4 amu. The isotopes found in nature have
either a mass of 62amu or 66amu. Determine the isotopic distribution.