Chapter 4 Chemical Equations & Stoichiometry Chemical reactions are best described using equations which tells us what compounds we started with (reactants), what we did to them (reaction conditions) and what compounds we ended up with (products). e.g. hυ C 2 H 6 + Br 2 C 2 H 5 Br + HBr CHEMICAL EQUATIONS SHOULD ALWAYS BE BALANCED. If an equation is unbalanced, correct it using the following guidelines: 1. Write the equation using one molecule of each type. 2. Check that each molecule has the right formula. 3. Balance one element at a time. 4. Balance polyatomic ions as a group. 5. Balance large molecules before small molecules. 6. Balance oxygen last (unless there is another element in more molecules; if so, balance that element last). 7. Ensure that the coefficients are the smallest possible whole numbers ( simplest whole number ratio ). 8. Add states of matter. Fe + O 2 Fe 2 O 3 (rust formation) C 3 H 8 + O 2 CO 2 + H 2 O (burning propane) C 8 H 18 + O 2 CO 2 + H 2 O (burning gasoline)
Chemical equations can be added (or subtracted). Generally, this is done to describe the overall result of a multi-step reaction. e.g. Fe 2 O 3 + 3 C 2 Fe + 3 CO Fe 2 O 3 + 3 CO 2 Fe + 3 CO 2 Chemical equations can be multiplied (or divided) by any number. Every coefficient gets multiplied by this number. e.g. 2 [Mg + 2 AgNO 3 2 Ag + Mg(NO 3 ) 2 ] Stoichiometry When we perform a chemical reaction, we often want to know how much product we expect to make. We can measure our success by comparing this number to how much product we actually make. In order to calculate how much product we should make, we need to know the ratio in which the reactants react and the product is formed. This is called stoichiometry. Recall the law of conservation of mass (chapter 2): Matter is neither created nor destroyed in a chemical reaction. Thus, for every atom of an element in the reactants, there must be an atom of the same element in the products. We can use this knowledge to calculate how much product can be made from a given amount of reactant (or vice versa).
e.g. CH 4 + 2 O 2 CO 2 + 2 H 2 O mole of CH 4 reacts with moles of O 2 to make mole of CO 2 and moles of H 2 O If we burn 100 grams of CH 4 with enough oxygen, what mass of water is produced (assuming complete reaction)? CH 4 + 2 O 2 CO 2 + 2 H 2 O *** Stoichiometric calculations must always use moles (not just masses). This is why it is essential to have balanced reaction equations***
e.g. Iron(III) oxide reacts with carbon to give iron and carbon dioxide. How many grams of carbon are required to produce 12.5 g iron? Limiting Reactants It s rare for a chemist to mix reactants in the perfect stoichiometric amounts. More often, one or more reactant will be added in excess in order to make sure that the most important/expensive reactant gets completely consumed. This one reactant is called the limiting reactant (or limiting reagent). The limiting reactant is not necessarily the one with the lowest mass or even the least moles. To find the limiting reactant, divide the number of moles of each reactant by the number of moles required. The reactant with the lowest number of ratioadjusted moles is the limiting reactant.
The limiting reactant is the only one used to calculate the amount of product formed from a reaction. e.g. Some cutting tools are protected by coating them with a thin layer of TiB 2 (a very hard compound). TiB 2 is formed by reacting titanium(iv) chloride, boron trichloride and hydrogen gas at 1000 C. The only other product of this reaction is hydrogen chloride. (a) Write a balanced chemical equation for this reaction. (b) If 200 g of each reactant were present in the reaction chamber, what mass of TiB 2 would be formed? (c) What mass of each reactant would remain once the reaction was complete?
Percent Yield Percent yield is calculated by comparing the theoretical yield (the maximum possible amount of product) and the actual yield (how much product was actually obtained): percent yield = actual yield x 100 % theoretical yield It is always necessary to use moles and stoichiometry to calculate percent yield. You cannot simply divide the mass of product(s) by the mass of reactant(s).
e.g. Propane (C 3 H 8 ) burns in oxygen to produce carbon dioxide and water. Calculate the mass of carbon dioxide produced if the reaction of 45.0 g of propane and sufficient oxygen has a 60.0% yield.
Quantitative Analysis Quantitative analysis is the identification of an unknown substance by subjecting it to chemical reactions and analyzing the resulting products. (What are they? How much of each was made?) Generally, we must already know which elements the unknown contains in order to choose the best reactions. Quantitative analysis is essentially stoichiometry backwards. As such, it is necessary to have a balanced chemical equation and work with moles and mole ratios. e.g. 5 L of a potassium iodide solution is reacted with 500 ml of a lead(ii) nitrate solution to produce lead(ii) iodide as a yellow solid with a mass of 27.14 g. What was the molarity of the initial lead(ii) nitrate solution? (Assume lead(ii) nitrate was the limiting reactant.)
e.g. An unknown organic compound has the formula C x H y O z. Upon complete combustion of 3.16 g C x H y O z with excess oxygen gas, 8.79 g carbon dioxide and 1.44 g water are isolated. What is the empirical formula of C x H y O z?
Quantitative analysis can also be used to calculate the percent composition of a known compound in a contaminated sample. For this type of question, make sure to distinguish clearly between the mass of pure compound (i.e. how much compound actually reacts) and the mass of the impure sample e.g. At high temperatures, sodium bicarbonate is converted quantitatively to sodium carbonate, carbon dioxide and water. A 1.7184 g sample of impure sodium bicarbonate gives off 0.196 g carbon dioxide upon heating. Assuming the reaction went to completion, what was the mass percent of sodium bicarbonate in the original sample? (Kotz, p.168)
Important Concepts from Chapter 4 balancing chemical equations calculations o stoichiometry o limiting reactants o percent yield o quantitative analysis