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Solving. Chemical Equation Word Problems. The "Mole". Definition:. One mole is the amount of any pure substance containing the same number of chemical units (atoms, molecules, ions, electrons, etc.) as there are atoms in exactly 12 grams of carbon-12. Exactly 12 grams of carbon-12 contains:
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Solving Chemical Equation Word Problems
The "Mole" Definition: One moleis the amount of any pure substance containing the same number of chemical units (atoms, molecules, ions, electrons, etc.) as there are atoms in exactly 12 grams of carbon-12. Exactly 12 grams of carbon-12 contains: 6.02 x 1023 atoms of carbon (Note: For atomic weights, we will round periodic atomic masses to the nearest whole number. For example, H = 1, Li = 7, O = 16.)
To find a mole of a substance (element or compound) use this two-step process: Step 1: Find the atomic mass of the element or the molecular weight of the compound (add all the masses of all the atoms in the compound). Step 2: Add “grams” as the unit to the number you found in step 1. Example 1: Find one mole of lithium Step 1. The atomic mass of lithium is 7 Step 2. One mole of lithium is 7 grams
Example 2: Find one mole of carbon dioxide, CO2 Step 1. The molecular weight of CO2 is 44 1 C 1 x 12.0 = 12 2 O 2 x 16.0 = 32 44 Step 2. One mole of CO2 is 44 grams
Moles to Grams Grams to Moles How many moles in 15 grams of lithium? 7 g • Find grams in one mole of the substance: • Set up a ratio: • Cross multiply and divide to find unknown: 1 mole = X 7 g 15 g 1 mole x 15 g = X x 7 g 7 g 7 g X = 2.14 mol
Moles to Grams Grams to Moles How many moles in 15 grams of lithium? • Find grams in one mole of the substance to set up conversion factor: 2. Set up the dimension: 3. Do the math: 1 mol = 7 g 15 g x 1 mol = 7 g 15 g x 1 mole = 2.14 mol 7 g
Moles to Grams Grams to Moles How many grams in 2.4 moles of sulfur? 32 g • Find grams in one mole of the substance: • Set up a ratio: • Cross multiply and divide to find unknown: 1 mole = 2.4 mol 32 g X 1 mole x X = 2.4 mol x 32 g X = 76.8 g 1 mole 1 mole
Solving Chemical Equation Word Problems • A six step process that can be used to solve just about any • chemical equation word problem that involves finding out • how much of one substance is needed to react with another • substance, or how much of a product is the result of • a chemical reaction. • This process requires the following six steps (skills): • 1. writing chemical formulas (we have done) • 2. finding molar masses (we have done) • 3. balancing chemical equations (we have done) • 4. using coefficients to find molar equivalents (new) • (you have done) 5. setting up and using simple math ratio & proportion • (you have done) 6. cross multiplying and dividing
Solving Chemical Equation Word Problems Use the (six) step by step process to solve the following problem: How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water?
Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Writing chemical formulas. Step 1: Identify the reactants and products and then write the chemical equation using the appropriate chemical symbols and formulas. Identify the reactants: nitrogen trihydride oxygen NH3 + O2 → Identify the products: nitrogen gas water N2 + H2O Write the chemical equation using appropriate chemical symbols and formulas. NH3 + O2 → N2 + H2O
Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Finding molar masses. Step 2: Before balancing the equation, use your periodic table to find the mass of one mole of each element or compound. Write the molar masses ABOVE the element or compound. One mole = 17 g 32g 28 g 18 g (molar masses) NH3 + O2 → N2 + H2O
Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Balancing chemical equations. Step 3: Balance the equation using correct coefficients. One mole =17 g 32g 28 g 18 g(molar masses) NH3 + O2 → N2 + H2O 432 6 Note: Coefficients will represent the number of MOLES of a substance!!!!
Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Using coefficients to find molar equivalents. Step 4: Multiply the molar masses by the coefficients (moles) to get the balanced molar mass equivalents. Write this number below the symbol of the element or compound. One mole =17 g 32g 28 g 18 g(molar masses) 4 NH3 + 3 O2 → 2 N2 + 6 H2O 68 g 96 g 56 g108 g
Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Setting up and using simple math ratio & proportion (fractions). Step 5: Using your problem statement and the balanced mass equivalents, set up a fractional ratio and proportion to solve for your unknown. A. Identify the unknown (the quantity you are trying to find = How much nitrogen gas in grams?) and the information that you do know (you have 10 grams of ammonia) and line them up right under the balanced mass equivalents for the chemicals that represent the same measurements. 4 NH3 + 3 O2 → 2 N2 + 6 H2O 68 g 96 g 56 g108 g 10 g X →
B. Since we only need two fractions to solve for one unknown, we can now set up a simple equation to solve. (NOTE: Since oxygen and water are not part of this particular problem, we need them to balance the equation, but do not need them for the final solution.) 68 g= 56 g 10 g X
Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Cross multiplying and dividing. Step 6: Cross multiply and divide to solve for the unknown (X) quantity. 68 g= 56 g 10 g X a. 68 g x X = 10 g x 56 g b. 68 X g = 560 g g c. 68 X g = 560 g g X = 8.2 g N2 68 g 68 g
Solving Chemical Equation Word Problems Practice problem # 1: Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to make water. If you start with 100 grams of hydrogen gas and all the oxygen that you need to completely combine with the hydrogen, then how much water will be produced? hydrogen gas + oxygen gas → water 1 mol = 2 g 32 g 18 g H2 + O2→ H2O 4 2 8 g + = 36 g 100 g X X = 450 g
Solving Chemical Equation Word Problems Practice problem # 2: You have 100 grams of sodium that you are going to combine with chlorine gas (diatomic) to make sodium chloride (salt). How much chlorine gas will you need to just exactly combine with the 100 grams of sodium? sodium + chlorine gas → sodium chloride 1 mol = 23 g 70 g 58 g Na + Cl2→ NaCl 2 2 46 g = 70 g 100 g X X = 152 g
Your turn! Complete and turn in the worksheet (Due next class).