• chemical stoichiometry: consider the quantity of materials consumed and produced in chemical reactions.

3.1 Counting by Weighing

• The most accurate method currently available for comparing the masses of atoms involves the use of the mass spectrometer.
  • Atoms or molecules are passed into a beam of high-speed electrons, which knock electrons off the atoms or molecules being analyzed and change them into positive ions.
  • An applied electric field then accelerates these ions into a magnetic field.
  • Because an accelerating ion produces its own magnetic field, and interaction with the applied magnetic field occurs, which tend to change the path of the ion.
  • The amount of path deflection for each ion depends on its mass — the most massive ions are deflected the smallest amount — which causes the ions to separate.
  • A comparison of the positions where the ions hit the detector plate gives very accurate values of their relative masses.
• Average atomic mass
  • Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon t be composed of only one type of atom with a mass of 12.01.

3.3 The Mole

• Mole: the number equal to the number of carbon atoms in exactly $\ce{12 g}$ of pure $\ce{^{12}C}$.

There are $\ce{10.0 g}$ sample of aluminum. Calculate the moles and number of atoms in the sample of $\ce{Al}$. number of moles: $\ce{10.0 g Al \times \frac{1 mol Al}{26.98 g Al} = 0.371 mol Al atoms}$ number of atoms: $\ce{0.371 mol Al \times \frac{6.022 \times 10^{23} atoms}{1 mol Al} = 2.33 \times 10^23 atoms}$

Calculate both the number of moles in a sample of cobalt (Co) containing $5.00 \times 10^{20}$ atoms and the mass of the sample. $\ce{5.00 \times 10^{20} atoms Co \times \frac{1 mol Co}{6.022 \times 10^{23} atoms Co} = 8.30 \times 10^{-4} mol Co}$ $\ce{8.30 \times 10^{-4} mol Co \times \frac{58.93 g Co}{1 mol Co} = 4.89 \times 10^{-2}g Co}$

3.4 Molar Mass

• The molar mass of a substance is the mass in grams of 1 mole of the compound.

$\ce{Mass of 1 mol C = 12.01 g}$ $\ce{Mass of 4 mol H = 4 \times 1.008 g}$ $\ce{Mass of 1 mol CH4 = 16.04 g}$

A. Calculate the molar mass of calcium carbonate. In 1 mole of calcium carbonate, there are 1 mole of $\ce{Ca^2+}$ ions and 1 mole of $\ce{CO3^2-}$ ions. Molar mass: $\ce{1 Ca^2+: 1 \times 40.08 g = 40.08 g}$ $\ce{1 CO3^2-: }$ $\ce{1 C: 1 \times 12.01 g = 12.01 g}$ $\ce{3 O: 3 \times 16.00 g = 48.00 g}$ $\ce{Mass of 1 mole of CaCO3 = 100.09 g}$

B. A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this sample? What is the mass of the $\ce{CO3^2-}$ ions present? $\ce{4.86 mol CaCo3 \times \frac{100.09 g CaCO3}{1 mol CaCO3} = 486 g CaCO3}$ The mass of 1 mole of $\ce{CO3^2-}$ ions is: $\ce{Mass of 1 mol CO3^2- = 1 \times 12.01 + 3 \times 16.00 = 60.01 g}$ Therefore, the mass of 4.86 moles of $\ce{CO3^2-}$ is $\ce{4.86 mol CO3^2- \times \frac{60.01 g CO3^2-}{1 mol CO3^2-} = 292 g CO3^2-}$

3.5 Learning to Solve Problems

• conceptual problem solving: solve problems in a flexible creative way based on understanding the fundamental ideas of chemistry.

1. We need to read the problem and decide on the final goal. Then we sort through the facts given, focusing on the key words and often drawing a diagram of the problem. In this part of the analysis we need to state the problem as simple and as visually as possible. We could summarize this entire process as “Where are we going?”
2. In order to reach our final goal, we need to decide where to start. For example, in a stoichiometry problem we always start with the chemical reaction. Then we ask a series of questions as we proceed, such as, “What are the reactants and products?” “What is the balanced equation?” “What are the amounts of the reactants?” and so on. Our understanding of the fundamental principles of chemistry will enable us to answer each of the these simple questions and eventually will lead us to the final solution. We might summarize this process as “How do we get there?”