Unit 6 Part 2 Mole Related Calculations
Several mole related calculations % composition Find the % of each element in a compound Empirical Formula Determine the simplest formula from mass or % composition data Molecular Formula Determine molecular formula from mass or % composition date Hydrates Recognize a hydrate; recognize the name of a hydrate Determine the hydrate formula from mass or % composition data
% Composition The percentage composition of a compound tells you the percent of the mass made up by each element in the compound. Steps for solving % composition problem 1. Determine molar mass of the compound. 2. Then find the % of each element in the compound by using the following formula % x 100%
Example: find the % of each element in Ca(NO 2 ) 2 Step 1 find molar mass 1 Ca 1 x 40.08 = 40.08 2 N 2 x 14.01 = 28.02 4 O 4 x 16.00 = 64.00 132.10 g Step 2 complete formula for each element %.. x 100 % = 30.34 % Ca If all is done correctly, then the percentages should total to be very close to 100%. %.. x 100 % = 21.21 % N %.. x 100 % = 48.45 % O
Empirical Formula A formula that gives the simplest whole-number ratio of the atoms of the elements in a chemical compound is called an empirical formula. The empirical formula simply tells you the ratio of the atoms in the compound. For glucose, C 6 H 12 O 6, the empirical formula is CH 2 O. (Notice that all of the subscripts in the formula could be reduced by the number 6.)
Steps for an Empirical Formula Problem 1. Convert the mass of each element to moles. (Keep significant digits) 2. Do the mole ratio (which means, divide each mole answer by the smallest mole number to get a whole number...double all if you get half a whole number) 3. The whole numbers become the subscripts in the formula.
What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? Step 1 - find moles of each element 53.73 g Fe 1 molfe =.9620 molfe 55.85 g Fe 46.27 g S 1 mols =.1437 mols 32.07 g S Step 2 do mole ratio (looking for whole #).9620 molfe = 1 Fe.1437 mols = 1.5 S.9620.9620 Step 3 write the formula This was an unusual case where you did not get a whole number relationshiip for the ratio, but found a half of a whole number. When that happens, you must double both subscripts to obtain the correct formula. Cannot have F 1 S 1.5, so double both to get F 2 S 3
Molecular Formulas The formula that gives the actual number of atoms of each element in a molecular compound is called the molecular formula. The molecular formula is a whole-number multiple of an empirical formula. You can determine the molecular formula by comparing the empirical formula mass to the molecular formula mass. The next shows a comparison between four molecular formulas and their empirical formula.
Look at the Examples Below Molecule Name Molecular Formula Empirical Formula Whole # Multiple Glucose C 6 H 12 O 6 CH 2 O 6 Tetrose C 4 H 8 O 4 CH 2 O 4 Acetic Acid C 2 H 4 O 2 CH 2 O 2 Formaldehyde CH 2 O CH 2 O 1 Notice a few things: 1. Each of the molecular formulas can be reduced to the same empirical formula. 2. The empirical formula in these examples happens to represent a real compound. 3. The empirical formula does not have to represent a real compound. 4. There is always a whole number that relates the molecular formula to an empirical formula. 5. The whole idea behind these problems is to determine the whole number by which to multiple the empirical formula in order to ascertain the molecular formula.
Steps for Solving Molecular Formula Problems 1. Determine the empirical formula (this can be either finding the empirical formula given in the problem or determining it from % composition data). 2. Find the empirical formula mass (use periodic table). 3. Divide the molecular formula mass (given in the problem) by the empirical formula mass (that you just determined in step 2) to get a whole number. 4. Adjust the empirical formula by the whole number. (In other words, multiply all of the subscripts in the empirical formula by the whole number to get the molecular formula.)
An organic compound is found to contain 92.25% carbon and 7.75% hydrogen. If the molecular mass is 78 g/mole, what is the molecular formula? (notice: have to follow empirical formula steps first) Step 1 of molecular formula find empirical formula Step 1 find moles of each element Step 2 of molecular formula - 92.25 g C 1 molc = 7.681 molc find empirical formula mass 12.01 g C 1 C 12.01 g 1 H 1.01 g 7.75 g H 1 molh = 7.67 molh 13.02 g 1.01 g H Step 3 of molecular formula - Step 2 do mole ratio divide molecular mass by 7.681 molc = 1 C 7.67 molh = 1 H empirical formula mass 7.67 7.67 78 g 13.02 g = 6 Whole # Step 3 write empirical formula Step 4 of molecular formula CH (CH) 6 = C 6 H 6
Hydrates Hydrated crystals are ionic compounds that have water molecules attached. That does not mean that the compounds are wet. They are actually very pretty crystals MgSO 4 7H 2 O is a white crystal. CuSO 4 5H 2 O is a blue crystal When naming hydrates, you will use the same prefixes as before. MgSO 4 7H 2 O -magnesium sulfate heptahydrate CuSO 4 5H 2 O copper II sulfate pentahydrate
Anhydrous Compounds The water can be driven off the hydrate by heating the compound. It is then known as the anhydrous compound (w/o water). Once the water is driven off, the compounds are no longer pretty crystals. They usually look like a white cake-like powder. MgSO 4 7H 2 O becomes MgSO 4 CuSO 4 5H 2 O becomes CuSO 4
Steps for Hydrate Problems 1. Find moles of anhydrous, moles of water 2. Do mole ratio (divide each mole answer by the smaller mole, looking for a whole #) 3. Whole # becomes coefficient for water (please note, that the whole # for the first part of the compound will always be 1 and the water will have a whole # of 1 or higher ) The following example happens to be #1 from worksheet 9.
A hydrated compound has 3.71 g Na 2 CO 3 and 6.29 g H 2 O. What Is the formula for this hydrate? Step 1 find moles anhydrous compound, find moles water 2 Na 2 x 22.99 g = 45.98 g 2 H 2 x 1.01 = 2.02 g 1 C 1 x 12.01 g = 12.01 g 1 O 1 x 16.00 = 16.00 g 3 O 3 x 16.00 g = 48.00 g 18.02 g 105.99 g 3.71 g Na 2 CO 3 1 mole Na 2 CO 3 = 0.0350 molna 2 CO 3 105.99 g Na 2 CO 3 6.29 g H 2 O 1 molh 2 O = 0.349 molh 2 O 18.02 g H 2 O Step 2 do mole ratio 0.0350 molna 2 CO 3 = 1 Na 2 CO 3 0.349 molh 2 O = 10 H 2 O 0.0350 0.0350 Step 3 use whole number to write formula Na 2 CO 3 10H 2 O