# Major chemistry laws. Mole and Avogadro s number. Calculating concentrations.

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1 Major chemistry laws. Mole and Avogadro s number. Calculating concentrations. Major chemistry laws Avogadro's Law Equal volumes of gases under identical temperature and pressure conditions will contain equal numbers of particles (atoms, ion, molecules, electrons, etc.). Boyle's Law At constant temperature, the volume of a confined gas is inversely proportional to the pressure to which it is subjected. PV = k Charles' Law At constant pressure, the volume of a confined gas is directly proportional to the absolute temperature. V = kt Combining Volumes Refer to Gay-Lussac's Law Conservation of Energy Energy can be neither created nor destroyed; the energy of the universe is constant. This is the First Law of Thermodynamics. Conservation of Mass Also known as Conservation of Matter. Matter can be neither created nor destroyed, though it can be rearranged. Mass remains constant in an ordinary chemical change. Dalton's Law The pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. Definite Composition A compound is composed of two or more elements chemically combined in a defined ratio by weight. Dulong & Petit's Law Most metals require 6.2 cal of heat in order to raise the temperature of 1 gram-atomic mass of the metal by 1 C. Faraday's Law The weight of any element liberated during electrolysis is proportional to the quantity of electricity passing through the cell and also to the equivalent weight of the element. First Law of Thermodynamics

2 Conservation of Energy. The total energy of the universe is constant and is neither created nor destroyed. Gay-Lussac's Law The ratio between the combining volumes of gases and the product (if gaseous) can be expressed in small whole numbers. Graham's Law The rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular mass. Henry's Law The solubility of a gas (unless it is highly soluble) is directly proportional to the pressure applied to the gas. Ideal Gas Law The state of an ideal gas is determined by its pressure, volume, and temperature according to the equation: PV = nrt where P is the absolute pressure V is the volume of the vessel n is the number of moles of gas R is the ideal gas constant T is the absolute temperature Multiple Proportions When elements combine, they do so in the ratio of small whole numbers. The mass of one element combines with a fixed mass of another element according to this ratio. Periodic Law The chemical properties of the elements vary periodically according to their atomic numbers. Second Law of Thermodynamics Entropy increases over time. Another way of stating this law is to say that heat cannot flow, on its own, from an area of cold to an area of hot.

3 Molecules and Moles A molecule is a combination of two or more atoms that are held together by covalent bonds. A molecule is the smallest unit of a compound that still displays the properties associated with that compound. Molecules may contain two atoms of the same element, such as O 2 and H 2, or they may consist of two or more different atoms, such as CCl 4 and H 2 O. In the study of chemistry, molecules are usually discussed in terms of their molecular weights and moles. Ionic compounds, such as NaCl and KBr, do not form true molecules. In their solid state, these substances form a three-dimensional array of charged particles. In such a case, molecular weight has no meaning, so the term formula weight is used instead. Molecular Weight and Formula Weight The molecular weight of a molecule is calculated by adding the atomic weights (in atomic mass units) of the atoms in the molecule. The formula weight of an ionic compound is calculated by adding its atomic weights according to its empirical formula. The Mole A mole is defined as the quantity of a substance that has the same number of particles as are found in grams of carbon-12. This number, Avogadro's number, is 6.022x The mass in grams of one mole of a compound is equal to the molecular weight of the compound in atomic mass units. One mole of a compound contains 6.022x10 23 molecules of the compound. The mass of 1 mole of a compound is called its molar weight or molar mass. The units for molar weight or molar mass are grams per mole. Here is the formula to determing the number of moles of a sample: mol = weight of sample (g) / molar weight (g/mol)

4 Exercises 1) How many H 2 O molecules are there in a snowflake weighing 1 mg? Snowflakes are made of water, H 2 O. First, we need to find the mass of one mole (Avogadro's number or x ) of H 2 O molecules. To obtain this number, first look up the atomic masses for H and O from the Periodic Table. There are two hydrogen atoms and one oxygen atom in water, so the mass of one mole of water is: 1.01 g x g x 1 = g This relation is used to determine how many molecules of water there are in 1 gram: number of molecules per gram = x H2O molecules / g number of molecules per gram = 3.34 x g Next, convert grams to milligrams. There are 1000 milligrams in a gram, so move the decimal point three places to the right or add three to the superscript in exponential notation: number of molecules in 1 milligram = 3.34 x x H 2 O molecules 2) Calculate the mass in grams of a single carbon (C) atom. To calculate the mass of a single atom, first look up the atomic mass for carbon from the Periodic Table. This number, 12.01, is the mass of one mole of carbon. One mole of carbon is x atoms of carbon (Avogadro's number). This relation is then used to 'convert' a carbon atom to grams: mass C atom = 1 C atom x g / x C atoms = x g x g

5 3) Calculate the mass in grams of 2.5 x 10 9 H 2 O molecules. First, we need to find the mass of one mole (Avogadro's number or x ) of H 2 O molecules. To obtain this number, first look up the atomic masses for H and O from the Periodic Table. There are two hydrogen atoms and one oxygen atom in water, so the mass of one mole of water is: 1.01 g x g x 1 = g This relation is used to complete the problem: mass = 2.5 x 10 9 H 2 O molecules x g / x H 2 O molecules mass = 7.5 x g 7.5 x g 4) The mineral cassiterite is a compound of tin and oxygen. Chemical analysis of cassiterite shows that the mass percentages of tin and oxygen are 78.8 and 21.2, respectively. Determine the formula of this compound. We want to find the number of moles of each element in order to determine the ratios of the elements and the formula. To make the calculation easy (i.e., let the percentages convert directly to grams), let's assume we have 100 g of cassiterite. In a 100 gram sample, there are 78.8 g Sn and 21.2 g O. Now, look up the atomic masses for the elements from the Periodic Table. The atomic masses are found to be: Sn is O is The atomic masses provide a moles per gram conversion factor. Using the conversion factor, we can calculate the moles of each element: moles Sn = 78.8 g Sn x 1 mol Sn / g Sn = mol Sn moles O = 21.2 g O x 1 mol O / g O = 1.33 mol O The numbers of moles of each element are in the same ratio as the number of atoms Sn and O in cassiterite. To find the simplest whole number ratio, divide each number by the smallest number of moles: Sn: / = 1.00 O: 1.33 / = 2.00

6 The ratios indicate that there is one tin atom for every two oxygen atoms. Thus, the simplest formula of cassiterite is SnO 2. SnO 2 5) Vitamin C contains three elements: carbon, hydrogen, and oxygen. Analysis of pure vitamin C indicates that the elements are present in the following mass percentages: C = 40.9 H = 4.58 O = 54.5 Use the data to determine the simplest formula for vitamin C. We want to find the number of moles of each element in order to determine the ratios of the elements and the formula. To make the calculation easy (i.e., let the percentages convert directly to grams), let's assume we have 100 g of vitamin C. If you are given mass percentages, always work with a hypothetical 100 gram sample. In a 100 gram sample, there are 40.9 g C, 4.58 g H, and 54.5 g O. Now, look up the atomic masses for the elements from the Periodic Table. The atomic masses are found to be: H is 1.01 C is O is The atomic masses provide a moles per gram conversion factor. Using the conversion factor, we can calculate the moles of each element: moles C = 40.9 g C x 1 mol C / g C = 3.41 mol C moles H = 4.58 g H x 1 mol H / 1.01 g H = 4.53 mol H moles O = 54.5 g O x 1 mol O / g O = 3.41 mol O The numbers of moles of each element are in the same ratio as the number of atoms C, H, and O in vitamin C. To find the simplest whole number ratio, divide each number by the smallest number of moles: C: 3.41 / 3.41 = 1.00 H: 4.53 / 3.41 = 1.33 O: 3.41 / 3.41 = 1.00 The ratios indicate that for every one carbon atom there is one oxygen atom. Also, there are 1.33 = 4/3 hydrogen atoms. (Note: converting the decimal to a fraction is a matter of practice! You know the elements must be present in whole number ratios, so look for common fractions and become familiar with the decimal equivalents for fractions so you can recognize them.) Another way to express the atom ratio is to write it as 1 C : 4/3 H : 1 O. Multiply by

7 three to obtain the smallest whole-number ratio, which is 3 C: 4 H : 3 O. Thus, the simplest formula of vitamin C is C 3 H 4 O 3. C 3 H 4 O 3 Concentrations The concentration of a chemical solution refers to the amount of solute that is dissolved in a solvent. Once you have identified the solute and solvent in a solution, you are ready to determine its concentration. Concentration may be expressed several different ways, using percent composition by mass, mole fraction, molarity, molality, or normality. Percent Composition by Mass (%) This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100. Example: Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt. : 20 g NaCl / 100 g solution x 100 = 20% NaCl solution Molarity (M) Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!). Example: What is the molarity of a solution made when water is added to 11 g CaCl 2 to make 100 cm 3 of solution? : 11 g CaCl 2 / (110 g CaCl 2 / mol CaCl 2 ) = 0.10 mol CaCl cm 3 x 1 dm 3 / 1000 cm 3 = 0.10 dm 3 molarity = 0.10 mol / 0.10 dm 3 molarity = 1.0 mol/ dm 3 (M)

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