The Mole Concept-Study Guide PDF

Study Guide Notes The chemical changes observed in any reaction involve the rearrangement of billions of atoms. It is impractical to try to count or visualize all these atoms, but scientists need some way to refer to the entire quantity. They also need a way to compare these numbers and relate them...

Study Guide Notes The chemical changes observed in any reaction involve the rearrangement of billions of atoms. It is impractical to try to count or visualize all these atoms, but scientists need some way to refer to the entire quantity. They also need a way to compare these numbers and relate them to the weights of the substances, which they can measure and observe. The solution is the concept of the mole, which is very important in quantitative chemistry. Avogadro’s Number Amadeo Avogadro first proposed that the volume of a gas at a given pressure and temperature is proportional to the number of atoms or molecules, regardless of the type of gas. Although he did not determine the exact proportion, he is credited for the idea. Avogadro’s number is a proportion that relates molar mass on an atomic scale to physical mass on a human scale. Avogadro’s number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to 6.02×1023 and is expressed as the symbol NA. Avogadro’s number is a similar concept to that of a dozen or a gross. A dozen molecules is 12 molecules. A gross of molecules is 144 molecules. Avogadro’s number is 6.02×1023molecules. With Avogadro’s number, scientists can discuss and compare very large numbers, which is useful because substances in everyday quantities contain very large numbers of atoms and molecules. The Mole The mole (abbreviated mol) is the SI measure of quantity of a “chemical entity,” such as atoms, electrons, or protons. It is defined as the amount of a substance that contains as many particles as there are atoms in 12 grams of pure carbon- 12. So, 1 mol contains 6.02×1023 elementary entities of the substance. Chemical Computations with Avogadro’s Number and the Mole Avogadro’s number is fundamental to understanding both the makeup of molecules and their interactions and combinations. For example, since one atom of oxygen will combine with two atoms of hydrogen to create one molecule of water (H2O), one mole of oxygen (6.02×1023 of O atoms) will combine with two moles of hydrogen (2 × 6.02×1023 of H atoms) to make one mole of H2O. Another property of Avogadro’s number is that the mass of one mole of a substance is equal to that substance’s molecular weight. For example, the mean molecular weight of water is 18.015 atomic mass units (amu), so one mole of water weight 18.015 grams. This property simplifies many chemical computations. If you have 1.25 grams of a molecule with molecular weight of 134.1 g/mol, how many moles of that molecule do you have? 1.25g ÷ 134.1 g/mol = 0.0093 mol Measuring Mass in Chemistry Chemists can measure a quantity of matter using mass, but in chemical reactions it is often important to consider the number of atoms of each element present in each sample. Even the smallest quantity of a substance will contain billions of atoms, so chemists generally use the mole as the unit for the amount of substance. One mole (abbreviated mol) is equal to the number of atoms in 12 grams of carbon-12; this number is referred to as Avogadro’s number and has been measured as approximately 6.02 x 1023. In other words, a mole is the amount of substance that contains as many entities (atoms, or other particles) as there are atoms in 12 grams of pure carbon-12. amu vs. g/mol Each ion, or atom, has a particular mass; similarly, each mole of a given pure substance also has a definite mass. The mass of one mole of atoms of a pure element in grams is equivalent to the atomic mass of that element in atomic mass units (amu) or in grams per mole (g/mol). Although mass can be expressed as both amu and g/mol, g/mol is the most useful system of units for laboratory chemistry. Calculating Molar Mass Molar mass is the mass of a given substance divided by the amount of that substance, measured in g/mol. For example, the atomic mass of titanium is 47.88 amu or 47.88 g/mol. In 47.88 grams of titanium, there is one mole, or 6.02x1023 titanium atoms. The characteristic molar mass of an element is simply the atomic mass in g/mol. However, molar mass can also be calculated by multiplying the atomic mass in amu by the molar mass constant (1 g/mol). To calculate the molar mass of a compound with multiple atoms, sum all the atomic mass of the constituent atoms. For example, the molar mass of NaCl can be calculated for finding the atomic mass of sodium (22.99 g/mol) and the atomic mass of chlorine (35.45 g/mol) and combining them. The molar mass of NaCl is 58.44 g/mol. Determining the Molar Mass of a Compound Here’s an example: In a compound of NaOH, the molar mass of Na alone is 23 g/mol, the molar mass of O is 16 g/mol, and H is 1 g/mol. What is the molar mass of NaOH? Na + O + H = NaOH 23 g/mol + 16 g/mol + 1 g/mol = 40 g/mol The molar mass of the compound NaOH is 40 g/mol. Converting Mass to Number of Moles Here’s an example: How many moles of NaOH are present in 90 g of NaOH? Since the molar mass of NaOH is 40 g/mol, we can divide the 90 g of NaOH by the molar mass (40 g/mol) to find the moles of NaOH. 90 g ÷ 40 g/mol = 2.25 mol Mass-Mole-# of particle Conversion Map  Percent Composition (by mass) Percent composition indicates the relative amounts of each element in a compound. For each element: Here’s an example: Bicarbonate of soda (sodium hydrogen carbonate) is used in many commercial preparations. Its formula is NaHCO3. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate. When doing mass percent calculations, it is always a good idea to check to make sure your mass percents add up to 100% (helps catch math errors): 27.38 + 1.19 + 14.29 + 57.14 = 100.00 Empirical Formula An empirical formula represents the lowest whole-number ratio of elements in a compound. Here’s how to find an empirical formula when given percent composition: 1. Assume that you have 100 g of the unknown compound. The beauty of this little trick is that you conveniently gift yourself with the same number of grams of each elemental component as its contribution to the percent composition. For example, if you assume that you have 100 g of a compound composed of 60.3% magnesium and 39.7% oxygen, you know that you have 60.3 g of magnesium and 39.7 g of oxygen. (The only time you don’t do this is if the problem specifically gives you the masses of each element present in the unknown compound.) 2. Convert the masses from Step 1 into moles using the molar mass. 3. Determine which element has the smallest mole value. Then divide all the mole values you calculated in Step 2 by this smallest value. This division yields the mole ratios of the elements of the compound. 4. If any of your mole ratios aren’t whole numbers, multiply all numbers by the smallest possible factor that produces whole-number mole ratios for all the elements. For example, if you have 1 nitrogen atom for every 0.5 oxygen atoms in a compound, the empirical formula is not N1O0.5. Such a formula casually suggests that an oxygen atom has been split, something that would create a small-scale nuclear explosion. Though impressive sounding, this scenario is almost certainly false. Far more likely is that the atoms of nitrogen and oxygen are combining in a 1 : 0.5 ratio but do so in a larger but equivalent ratio of 2 : 1. The empirical formula is thus N2O. Because the original percent composition data is typically experimental, expect to see a bit of error in the numbers. For example, 2.03 is probably within experimental error of 2, 2.99 is probably 3, and so on. 5. Write the empirical formula by attaching these whole-number mole ratios as subscripts to the chemical symbol of each element. Order the elements according to the general rules for naming ionic and molecular compounds. Here’s an example: What is the empirical formula of a substance that is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass? For the sake of simplicity, assume that you have a total of 100 g of this mystery compound. Therefore, you have 40.0 g of carbon, 6.7 g of hydrogen, and 53.3 g of oxygen. Convert each of these masses to moles by using the gram atomic masses of C, H, and O: Notice that the carbon and oxygen mole numbers are the same, so you know the ratio of these two elements is 1:1 within the compound. Next, divide all the mole numbers by the smallest among them, which is 3.33. This division yields The compound has the empirical formula CH2O. The actual number of atoms within each particle of the compound is some multiple of the numbers expressed in this formula. How to Find Molecular Formulas from Empirical Formulas To determine a molecular formula, you must know the molecular (formula) mass of the compound as well as the empirical formula (or enough information to calculate it yourself from the mass or percent composition of each element in the compound). With these tools in hand, calculating the molecular formula involves three steps:  1. Calculate the empirical formula mass. 2. Divide the gram molecular mass by the empirical formula mass. 3. Multiply each of the subscripts within the empirical formula by the number calculated in Step 2. Here‘s an example: What is the molecular formula of a compound that has a gram molecular mass of 34 g/mol and the empirical formula HO? 1. Calculate the empirical formula mass. You determine this number by finding the mass of HO (1 hydrogen atom and 1 oxygen atom). So, the empirical formula mass is 17.01 g/mol. 2. Divide the gram molecular mass by the empirical formula mass. Dividing the gram molecular mass by this value yields the following: 3. Multiply each of the subscripts within the empirical formula by the number calculated in Step 2. Multiplying the subscripts within the empirical formula by this number gives you the molecular formula H2O2. This formula corresponds to the compound hydrogen peroxide.  Study Guide Questions 1 What is the molar mass of H2CO3. (Atomic mass: C = 12.01, H = 1.01, O = 15.99) A 23.00 g/mol B 24.00 g/mol C 61.00 g/mol D 62.00 g/mol 2 What is the molar mass of CO2? F 22.0 g/mol G 28.0 g/mol H 44.0 g/mol J 56.0 g/mol 22 3 What mass of sulfur would have precisely 4.7x10 atoms of sulfur? A 6.02 g sulfur B 4.7 g sulfur C 2.5 g sulfur D 1.3 g sulfur 4 How many hydrogen atoms are in 2 mol of H2O F 6.02x1023 G 1.20x1024 H 2.41x1024 J 4.82x1024 5 How many molecules of methane gas (CH4) are in 32.1 grams of methane? A 4.52x1023molecules B 6.02x1023molecules C 1.20x1024molecules D 3.09x1024molecules 6 Vinegar, a common household item, is used in cleaning, cooking, baking and meat preservation. Vinegar contains acetic acid, CH3COOH, which gives vinegar its sour taste and pungent smell. What is the percent composition of carbon in acetic acid? F 40.0% G 41.4% H 20.0% J 33.3% 7 What is the percentage of sodium present in sodium hydrogen carbonate, NaHCO3? A 27.38% B 33.82% C 44.23% D 84.00% 8 Iron forms two different oxides: iron(II) oxide, in which iron ions carry a +2 charge; and iron(III) oxide, in which iron carries a +3 charge. What is the difference in the percent of iron in these two compounds? 9 What is the empirical formula for the compound that has 46 grams of sodium, 64 grams of sulfur, and 48 grams of oxygen? A Na2S2O3 B Na3S2O2 C Na2S2O2 D Na3S3O3 10 A compound is analyzed and determined to be 30.4% nitrogen and 69.6% oxygen. What is the empirical formula of the compound? F NO G NO2 H N2O J N2O2 11 An unknown organic molecule with empirical formula CxHy is analyzed through combustion and determined to be 79.85% carbon. What is the value of the "Y" subscript in the empirical formula? 12 What is the molecular formula of a compound whose molar mass is 88.0 and whose percent composition is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen? F C2H4O G C3H4O3 H C4H8O2 J C5H12O 13 A compound containing sodium, chlorine, and oxygen is 25.42 % sodium by mass. A 3.25 g sample gives 4.33x1022 atoms of oxygen. What is the empirical formula? 14 What is the mass percentage of water in the hydrate CuSO4∙5H2O? 15 Zinc chloride, ZnCl2, is 52.02 % chlorine by mass. a) What mass of chlorine is contained in 80.3 g ZnCl2? b) How many moles of Cl is this?