Read Section 5.4 before viewing the slide show.

Unit 19 Calculations with the Mole Conversion Factor Approach (5.4) Mole-to-Mass and Mass-to-Mole Conversions (5.4) Mole and Mass Relationships in Chemical Equations (5.4)

Conversion Factor Approach A common approach for calculating a wide range of quantities is called the conversion factor approach. Its premises are: Multiplying any quantity by the number “1” returns the same quantity A factor which has a numerator equal to its denominator is equal to “1” A starting quantity can be multiplied by any number of conversion factors to simply change the units on the original quantity A simple example Convert 15 feet to yards (you can do this in your head – this just shows the approach) The conversion factor is 1 yd = 3 feet. The factor that has or is equal to “1” since the numerator and denominator are the same in each factor. Thus, our initial measured quantity may be multiplied by either one without changing the measured length. The calculation sets up as: The conversion factor is written in the orientation that cancels the original unit – feet – leaving yd for the final answer.

Mole-to-Mass and Mass-to-Mole Conversions From the previous Unit, we know that we can identify one mole of a material with Avogadro’s number of particles (6.02 x particles) which also a mass equal to the formula mass, molecular mass, or molar mass (all the same number) in grams. The mole then provides a convenient set of conversion factors that can be used in working problems. Consider NH 3 : 1 mol NH 3 = 6.02 x NH 3 molecules = 17 g NH 3 Example problems: How many grams of NH 3 are in 12.5 mol of NH 3 ? How many mol of NH 3 are in 145 g of NH 3 ? How many NH 3 molecules are in 35.0 g of NH 3 ? Desired Quantity Given Quantity Conversion Factor

Mole and Mass Relationships in Chemical Equations The conversion factor approach may also be used to address mass and mole relationships across a chemical equation. Consider the equation: C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O (g) A question is posed that asks how many grams of CO 2 could be produced from 50.0-g of C 3 H 8 reacted with an excess of oxygen. The mass of CO 2 produced will be dependent on the number of grams of C 3 H 8 since there is plenty of oxygen. The coefficients give the mole relationships in the chemical reaction. For every 1 mol of C 3 H 8 burned 3 mol of CO 2 are produced Since the relationships between moles and mass are known through the molar masses, there is sufficient information given to solve the problem. The next slide illustrates a method of handling this sort of problem using the conversion factor approach

Application of the Conversion Factor Approach in Chemical Equations The question is: How many grams of CO 2 can be produced from completely reacting 50.0-g of C 3 H 8 ? Consider the following table. The information in the first row is from the specific problem being presented – the mass of C 3 H 8 given and the unknown mass of CO 2 desired. The second row simply reproduces the chemical equation for reference. The third and fourth rows give information related to the chemical equation given. The coefficients in the chemical equation give the mole ratios and the conversion to mass is done by multiplying the number of moles times the molar mass. All quantities in the shaded area represent the same amount of chemical reaction – they can be taken in any pair to form a conversion factor that is equal to “1”.

Application of the Conversion Factor Approach in Chemical Equations The question is: How many grams of CO 2 can be produced from completely reacting 50.0-g of C 3 H 8 ? Consider the following table. First, set up write down the quantity being sought and the given amount in the conversion factor setup as: To determine the conversion factor to be used, recognize that its denominator will have to be in g C 3 H 8 to cancel the measured quantity and its numerator will have to be in g CO 2 to come up with the desired quantity. Those numbers are taken from the shaded area below the equation.

Application of the Conversion Factor Approach in Chemical Equations This approach is easily used to solve other problems related to the balanced chemical equation. Examples: How many g of O 2 will react with 75 g of C 3 H 8 ? How many moles of water will be formed in carrying out this reaction with 350 g of O 2 ?