Mole Stoichiometry Lab Report

Mole stoichiometry is at the cornerstone of quantitative Chemistry and frequently appears in many topics.

4 keys formulas:
1. no. of moles =

2. no. of moles = concentration in mol/(dm^3 ) x volume of solution in dm3

3. no. of moles =

4. no. of moles =

Stoichiometry is related to the whole number ratios (i) between atoms in chemical formulas and (ii) between reactants and products in a chemical equation.

Example 1
A mixture containing only aluminium oxide, Al2O3, and copper (I) oxide, Cu2O, weighs 2.02 g. When heated under a stream of hydrogen, only Cu2O is reduced to metallic copper…show more content…
of moles of Cu2O = no. of moles of O = 0.00875 mol (critical step)

mass of Cu2O = 0.00875 x (63.5 x 2 + 16.0) = 1.25125 g

percentage by mass of Cu2O = (1.25125 )/2.02 x 100 % = 61.9 %

Common mistake: no. of moles of (Al2O3 + Cu2O) =

Comment: no. of moles =

Example 2
0.144 g of a binary compound of aluminium and carbon reacts with an excess of water to produce a gas. This gas burns completely in oxygen to form water and 72 cm3 of carbon dioxide only. All measurements are taken at room temperature and pressure. Determine the empirical formula of the binary compound. (1 mole of gas occupies 24 dm3 of volume at r.t.p)
A Al2C3 B Al3C4 C Al4C3 D Al5C3
Solution
no. of moles of CO2 = 72/(24 000) = 0.003 mol

no. of moles of C = no. of moles of CO2 (from the formula, CO2, and the = 0.003 mol equation, C + O2  CO2) mass of C = 0.003 x 12.0 = 0.036 g

mass of Al = 0.144 – 0.036 = 0.108

aluminium carbon mass (g) 0.108 0.036 molar mass 27.0 12.0

(mass )/(molar mass) = no. of moles 0.108/27.0 = 0.004
0.036/12.0 = 0.003 ratio 4 3

Hence, the empirical formula of X is Al4C3.

Example…show more content…
of moles 5.0867 x 10-3 5.0682 x 10-2 5.1978 x 10-2 ratio 1 10 10

Hence, the empirical formula of Y is FeC10H10.

Calculations Involving Gases

Avogadro’s Law states that equal volumes of any gases, under the same conditions of temperature and pressure, contain the same number of particles (atoms or molecules). Molar volume of a gas at s.t.p (0oC and 1 atmosphere) is 22.4 dm3. Molar volume of gas at r.t.p (25oC and 1 atmosphere) is 24.0 dm3.
Volume of gas = molar volume x number of moles
The molecular formula of gaseous hydrocarbons (CxHy) and organic compounds (CxHyOz) can be determined by combustion in excess oxygen to form carbon dioxide and water, using the following equations:
CxHy + (x+ y/4) O2  x CO2 + y/2 H2O
CxHyOz + (x+ y/4- z/2) O2  x CO2 + y/2 H2O

Example 4

When 10 cm3 of a gaseous hydrocarbon was combusted in excess oxygen in an enclosed vessel, the volume of gas (measured at 298K) was reduced by 25 cm3. The addition of excess NaOH (aq) caused a further reduction in gas volume of 40 cm3 (measured at 298 K). The pressure in the vessel was maintained constant at 1 atm throughout the measurements. Find the molecular formula of the