Determining Percent Composition and Empirical Formulas of Compounds, Lecture notes of Chemistry

Download Determining Percent Composition and Empirical Formulas of Compounds and more Lecture notes Chemistry in PDF only on Docsity! Authored by Darren Rigby This work is licensed under a Creative Commons Attribution 4.0 International License Chemistry 0861 Learning Centre Percent Composition One way to identify an unknown compound is to determine the relative mass of the different elements in its make-up — its percent composition by mass. We can use the ideas of moles and molar mass to identify the compound. Before we look at that, it’s useful to see how a percent composition is calculated for a known compound. FINDING PERCENT COMPOSITION A percent composition is a breakdown of how much of the mass of any sample of a compound comes from each element in the compound. Example 1: What is the percent composition by mass of water? Solution: Water is H2O. To find percent composition, we’ll need the molar mass of the compound: 2 × H = 2 × 1.01 g⁄mol = 02.02 1 × O = 16.00 g⁄mol = 16.00 = 18.02 g⁄mol Now we divide this by the mass that each element contributed to the total mass. If an element contributes more than one atom to the molecule, use the mass of the total number of atoms for the element, not just one. H: 2.02 ÷ 18.02 = 0.11209… ≈ 11.2% O: 16.00 ÷ 18.02 = 0.88790… ≈ 88.8% So water is 11.2% hydrogen and 88.8% oxygen by mass. EMPIRICAL FORMULAS An empirical formula for a compound is a formula that tells you the ratio of atoms of the elements within it (the molar ratio), but not the exact number of atoms of each element in the molecule. What’s the difference? Consider NO2, nitrogen dioxide, which is produced in car exhaust, and N2O4, dinitrogen tetroxide, a component of rocket fuel. If I have a sample of each of them, then I know in both cases there are twice as many oxygen atoms in the sample as nitrogen atoms. They’re arranged into molecules in different ways, but the proportion of nitrogen to oxygen atoms is still 1 : 2. Both compounds have an empirical formula of NO2, even though they’re completely different chemicals. If we only know how much of a compound is nitrogen or oxygen by mass, it would be impossible to tell the two apart, but determining the empirical formula of a compound is still the important first step in this kind of analysis. This work is licensed under a Creative Commons Attribution 4.0 International License 2 Example 2: Determine the empirical formula of a compound that is 60.04% silicon and 39.96% nitrogen by mass. Solution: We start by assuming that we have 100 grams of this compound. (We’re working with a ratio, so any mass will work, and 100 g makes calculations very easy.) Then we figure out how many grams of silicon and oxygen would be in a 100-g sample. Last, we calculate how many moles of these elements are in it using their molar masses. 60.04% of 100 g = 60.04 g of Si 60.04 g Si × 1 mol Si 28.06 g Si = 2.1397… mol Si 39.96% of 100 g = 39.96 g of N 39.96 g N × 1 mol N 14.01 g N = 2.8522… mol N We still need to determine the ratio between the two. Divide each molar quantity by the smallest one you get. If a decimal results, hopefully you’ll recognize it as a fraction. (If not, try dividing it the other way.) 2.8522 ÷ 2.1397 = 1.3330… ≈ 11⁄3 = 4⁄3 If we divided them the other way, we get 0.7502, which is pretty close to 3⁄4. Either way, we recognize that the ratio is 3 Si : 4 N, so the empirical formula is Si3N4. We don’t know what this compound is exactly, yet. It could be Si3N4, or possibly something more complicated like Si6N8 or Si9N12…. We need more information to know for sure. A molar mass for the compound is (relatively) easy to measure and allows us to narrow the field down to one formula. A formula that says explicitly how many atoms of each element are in a compound is called a molecular formula. Example 3: (a) Determine the empirical formula of a compound that is 47.76% oxygen, 34.31% sodium and 17.93% carbon by mass. (b) Determine its molecular formula if its molar mass is 134.002 g⁄mol. Solution: (a) Assume we have 100 g of this compound. 47.76% of 100 g = 47.76 g of O 47.76 g × 1 mol O 16.00 g O = 2.985 mol O 34.31% of 100 g = 34.31 g of Na 34.31 g × 1 mol Na 22.99 g Na = 1.4923… mol Na 17.93% of 100 g = 17.93 g of C 17.93 g × 1 mol C 12.01 g C = 1.4929… mol C We divide: 2.985 ÷ 1.4923 ≈ 2, and 1.4929 ÷ 1.4923 ≈ 1. This means there are 2 oxygen atoms and one carbon atom for every sodium atom, so the empirical formula is O2NaC, or, putting the metal first and the non-metals in alphabetical order, NaCO2. (b) Now we have the molar mass. We can easily find out what the actual molecular formula is by dividing the molar mass of the empirical formula by the actual molar mass. Na + C + 2(O) = 22.99 + 12.01 + 2 × 16.00 = 67.00 g⁄mol 134.002 g⁄mol ÷ 67.00 ≈ 2