1. Chemistry XXI Unit 1-2 How do we distinguish substances? DISCUSSION Central goal: To apply the particulate model of matter to determine the composition and the number of particles in any given sample of a substance.

2. Chemistry XXI The chemical composition and structure of a substance determines its physical and chemical properties. Thus, one of the first steps in the analysis of a chemical substance is determining its composition: Let’s Think The elemental analysis of a chemical substance found in an explosive reveals the following composition: 27.04% Na, 16.48% N, 56.47% O. Determine its empirical formula.

3. Chemistry XXI Composition 1. In 100.0 g we have 27.04 g Na, 16.48 g N, 56.47 g O. 2. 27.04 g Na x 1 mole Na/ 22.99 g = 1.176 mol Na 16.48 g N x 1mole N/ 14.01 g = 1.176 mol N 56.47 g O x 1mole H/ 16.00 g = 3.529 mol O 3.Na 1.176 N 1.176 O 3.529  Na 1 N 1 O 3 NaNO 3 Sodium Nitrate

4. Chemistry XXI Amount Once we know the composition of a substance, we may be interested in quantifying the total amount of atoms or molecules that are present in certain mass of the material. M NANA The quantification relies on the use of two basic quantities: Molar mass (M)  Mass of one mole of substance; Avogadro’s Number (N A )  # of particles in one mole mass (m)    mol (n)Number of Particles (N)   

5. Chemistry XXI M NANA mass (m)    mol (n)Number of Particles (N)    Amount Keep in mind that, depending on the problem, there will be additional steps that need to be completed to figure out how much stuff we have. volume (V)    volume (V)   Gases PV = nRT

6. Chemistry XXI Let’s Think How many atoms of oxygen O are in 5.0 g of NaNO 3 ? M(NaNO 3 ) = 85.00 g/mol = 1.1 x 10 23 atoms of O M mass (m)    mol (n) NANA Number of Particles (N)   

7. Chemistry XXI Let’s Think What is the mass of a grain of salt (NaCl) formed by exactly 1x10 20 atoms ? M(NaCl) = 58.44 g/mol = 4.9 x 10 -3 g NaCl M mass (m)    mol (n) NANA Number of Particles (N)   

8. Chemistry XXI Amount vs. Concentration Many substances in our world, and in the laboratory, are dissolved in water. Thus, we need to learn how to quantify their amounts when they are in solution. An aqueous solution is characterized by its concentration: a)What is the difference between the total amount of solute in the solution and the concentration of the solution? b)Represent, using the particulate model of matter, a solution more concentrated than other but with a smaller amount of solute. Let’s Think

9. Chemistry XXI Amount vs. Concentration The amount of solute is a measure of the total quantity (mass, moles, particles) of that substance in the solution. The concentration of the solute is a measure of the quantity of substance (mass, moles, particles) per unit mass or volume. Solvent Solute

10. Chemistry XXI Molar Concentration The molar concentration (or molarity) of a substance in solution is defined as the number of moles of substance (n) per unit volume (V): Molarity = n / V Imagine that you want to prepare a solution of NaNO 3 containing 0.100 moles of the compound per every liter: 0.100 moles/L = 0.100 Molar = 0.100 M How many grams of NaNO 3 do you need to prepare 25.0 mL of the solution? Let’s Think

11. Chemistry XXI Concentration Molarity = n / V V = n / Molarity Given that M(NaNO 3 ) = 85.00 g/mol n = m/M(NaNO 3 ) = 0.85 / 85.00 = 0.01 mol V= 0.01 / 0.100 = 0.1 L = 100 mL If we want 0.0250 L of a 0.100 M solution : 0.100 M = n/0.025 L  n = 0.0025 mol We need 0.0025 x 85.00 = 0.212 g What should the total volume (in mL) of the solution be if you use 0.85 g of NaNO 3 to prepare it? Let’s Think