Chapter 6 Stoichiometry (Chemical Calculations) and the Mole Concept

Chapter 6: Stoichiometry (Chemical Calculations) and the Mole Concept THE MOLE CONCEPT - A mole describes a quantity of particles, 1 mole = 6.02 × 1023 particles - The number 6.02 × 1023 is also often referred to as Avogado’s number MOLAR MASS AND MOLES - When the formula mass of an ionic compound is determined by the addition of its component relative atomic masses and expressed in grams, it is called the molar mass. Ex: Find the molar mass of 𝐶𝑎𝐶𝑂3. 1 Ca = 40 1 C = 12 3 O = 48(3 × 16) 𝐶𝑎𝐶𝑂3 = 100.formula mass 100. g/mol is the molar mass - The molar mass is equivalent to the mass of 1 mole of that compound expressed in grams. MOLE RELATIONSHIPS - When dealing w/ elements- Moles of an element ×molar mass (atomic) = mass of the element - Mass of an element/molar mass (atomic) = moles of the compound - When dealing w/ compounds- Moles of a compound ×molar mass (molecular) = mass of the compound - Mass of a compound/molar mass (molecular) = moles of the compound - When dealing w/ the molecules of a compound - Moles of molecules × 6.02 × 1023 = number of molecules - Number of molecules/6.02 × 1023= moles of molecules - When dealing w/ the atoms of elements- Moles of atoms × 6.02 × 1023= number of atoms - Number of atoms/6.02 × 1023 = moles of atoms DENSITY AND MOLAR MASS - Since the density of a gas is usually given in grams/liter of gas at STP, we can use the molar volume to molar mass relationship to solve the following types of problems. Ex: Find the molar mass of a gas when the density is given as 1.25 grams/liter. - B/c we know that 1 mole of a gas occupies 22.4 liters at STP, we can solve this problem by multiplying the mass of 1 liter by 22.4 liters/mole. 1.25 𝑔 22.4 𝐿 × = 28.0 𝑔/𝑚𝑜𝑙 𝐿 1 𝑚𝑜𝑙 MASS -VOLUME RELATIONSHIPS Ex: How many liters of oxygen (STP) can you prepare from the decomposition of 42.6 grams of sodium chlorate? Step 1: Write the balanced equation for the reaction. 2 𝑁𝑎𝐶𝑙𝑂3 → 2 𝑁𝑎𝐶𝑙 + 3𝑂2 Step 2: Write the given quantity and the unknown quantity above the appropriate substances. 42.6 g xL 2 𝑁𝑎𝐶𝑙𝑂3 → 2 𝑁𝑎𝐶𝑙 + 3𝑂2 Step 3: Calculate the equation mass or volume of each substance that has something indicated above it, and write the results under the substances. 42.6 g xL 2 𝑁𝑎𝐶𝑙𝑂3 → 2 𝑁𝑎𝐶𝑙 + 3𝑂2 213 g 67.2 L (2×molar (3×22.4 L) Mass of 𝑁𝑎𝐶𝑙𝑂3) StEP 4: Form the proportion. 42.6 𝑔 𝑥𝐿 = 213 𝑔 67.2 𝐿 Step 5: Solve for x. x = 13.4 L 𝑂2 Step 4: Another method to proceed from Step 3 is called dimensional analysis or the factor-label method 42.6 𝑔 𝑁𝑎𝐶𝑙𝑂3 × 67.2 𝐿 𝑂2 213 𝑔 𝑁𝑎𝐶𝑙𝑂3 = 13.4 𝐿 𝑂2 - Another method of solving this problem is called the mole method. Step 3: Determine how many moles of substance are given. 106.5 𝑔 42.6 ÷ 1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙𝑂 = 0.400 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙𝑂3 3 The equation shows that 2 moles of 𝑁𝑎𝐶𝑙𝑂3 makes 3 moles 𝑂2 . Then 0.4 mole 𝑁𝑎𝐶𝑙𝑂3 will yield 3 𝑚𝑜𝑙 𝑂2 0.400 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙𝑂3 × 2 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙𝑂 = 0.600 𝑚𝑜𝑙 𝑂2 3 Step 4: Convert the moles of 𝑂2 to liters. 22.4 𝐿 𝑂2 0.600 𝑚𝑜𝑙 𝑂2 × = 13.4 𝐿 𝑂2 1 𝑚𝑜𝑙 𝑂2 MASS-MASS PROBLEMS Ex: What mass of oxygen can be obtained from heating 100. grams of potassium chlorate? Step 1: Write the balanced equation for the reaction. 2𝐾𝐶𝑙𝑂3 → 2𝐾𝐶𝑙 + 3𝑂2 Step 2: Write the given quantity and the unknown quantity above the appropriate substances. 100. g xg 2𝐾𝐶𝑙𝑂3 → 2𝐾𝐶𝑙 + 3𝑂2 Step 3: Calculate the equation mass for each substance that has something indicated above it, and write the results under the substances. 100. g xg 2𝐾𝐶𝑙𝑂3 → 2𝐾𝐶𝑙 + 3𝑂2 245 g 96.0 g Using the proportion method: Step 4: Form the proportion. 100.𝑔 𝑥𝑔 = 96.0 𝑔 245 𝑔 Step 5: Solve for x. x = 39.3 g of 𝑂2 Using the factor-label method: Step 4: The equation indicates that 245 g 𝐾𝐶𝑙𝑂3 yields 96 g of 𝑂2 . Therefore multiplying the given quantity by a factor made up of these two quantities arranged appropriately so that the units of the answer remain uncanceled will give the answer to the problem. 96 𝑔 𝑂 2 100. 𝑔 𝐾𝐶𝑙𝑂3 × 245 𝑔 𝐾𝐶𝑙𝑂 = 39.3 𝑔 𝑂2 3 Using the mole method: Step 3: Determine how many moles of substances are given. 1 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3 100. 𝑔 𝐾𝐶𝑙𝑂3 × = 0.815 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3 122.5 𝑔 𝐾𝐶𝑙𝑂3 The equation shows that 2 moles of 𝐾𝐶𝑙𝑂3 yields 3 moles of 𝑂2 , so 0.815 mole of 𝐾𝐶𝑙𝑂3 will yield 3 𝑚𝑜𝑙 𝑂2 0.815 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3 × 2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂 = 1.22 𝑚𝑜𝑙 𝑂2 3 Step 4: Convert the moles of 𝑂2 to grams of 𝑂2 . 32.0 𝑔 𝑂 1.22 𝑚𝑜𝑙 𝑂2 × 1 𝑚𝑜𝑙 𝑂2 = 39 𝑔 𝑂2 2 VOLUME-VOLUME PROBLEMS Ex: What volume of 𝑁𝐻3 is produced when 22.4 liters of nitrogen are made to combine w/ a sufficient quantity of hydrogen under the appropriate conditions? Step 1: Write the balanced equation for the reaction. 𝑁2 + 3𝐻2 → 2𝑁𝐻3 Step 2: Write the given quantity and the unknown quantity above the respective substances. 22.4 L xL 𝑁2 + 3𝐻2 → 2𝑁𝐻3 Using the proportion method: Step 3: Set up a proportion using the coefficient of the substances that have something indicated above them as denominators. 22.4 𝐿 𝑥𝐿 = 2𝐿 1𝐿 Step 4: Solve for x. x = 44.8 L Using the factor-label method: Step 3: The equation shows that 1 volume of 𝑁2 will yield 2 volumes of 𝑁𝐻3 . Therefore multiplying the given quantity by a factor made up of these 2 quantities appropriately arranged so that the units of the answer remain uncanceled will solve the problem. 2 𝐿 𝑁𝐻 22.4 𝐿 𝑁2 × 1 𝐿 𝑁 3 = 44.8 𝐿 𝑁𝐻3 2 Using the mole method: Step 3: The given quantity is converted to moles. 1 𝑚𝑜𝑙 𝑁 22.4 𝐿 𝑁2 × 22.4 𝐿 𝑁 2 = 1 𝑚𝑜𝑙 𝑁2 2 yield The equation shows that 1 mole of 𝑁2 will yield 2 moles of 𝑁𝐻3 . Therefore using this relationship will 1 𝑚𝑜𝑙 𝑁2 × 2 𝑚𝑜𝑙 𝑁𝐻3 1 𝑚𝑜𝑙 𝑁2 = 2 𝑚𝑜𝑙 𝑁𝐻3 Convert the moles of 𝑁𝐻3 to liters of 𝑁𝐻3 . 22.4 𝐿 𝑁𝐻 2 𝑚𝑜𝑙 𝑁𝐻3 × 1 𝑚𝑜𝑙 𝑁𝐻3 = 44.8 𝐿 𝑁𝐻3 3