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Section 9.3—Analysis of a Chemical Formula How can we determine a chemical formula?
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Percent Composition
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What is Percent Composition?
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Example #1 Example: A 8.5 g sample is composed of carbon and hydrogen. If 5.5 g of the sample is found to be carbon, what is the percent composition by mass of the sample?
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Example #1 If total = 8.5 g and C = 5.5 g then H =3.0 g - Example: A 8.5 g sample is composed of carbon and hydrogen. If 5.5 g of the sample is found to be carbon, what is the percent composition by mass of the sample?
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Let’s Practice #1 A sample is 57.0% by mass chlorine, how many grams of chlorine are present in a 27.5 g sample?
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Let’s Practice #1 27.5 g27.5 g 100 A sample is 57.0% by mass chlorine, how many grams of chlorine are present in a 27.5 g sample?
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Example #2 Percent composition can also be determined from a chemical formula Example: Find the percent composition, by mass, of CaCl 2
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Example #2 Percent composition can also be determined from a chemical formula total = 110.98 g Ca = 40.08 g Cl =70.90 g Ca Cl 1 2 40.08 g/mole 35.45 g/mole = 40.08 g/mole = 70.90 g/mole + 110.98 g/mole So for 1 mole: Example: Find the percent composition, by mass, of CaCl 2
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Let’s Practice #2 Find the percent composition, by mass, of NaNO 3
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Let’s Practice #2 Find the percent composition, by mass, of NaNO 3 Na N 1 1 22.99 g/mole 14.01 g/mole = 22.99 g/mole = 14.01 g/mole + 85.00 g/mole O316.00 g/mole = 48.00 g/mole
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Empirical Formulas
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What’s an Empirical Formula? Empirical – from data Empirical Formula – Chemical formula determined from lab data. Lowest possible ratio of atoms CH 2 is the lowest ratio (and empirical formula) of the molecule C 3 H 6
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Ratio of Atoms in a Molecule Subscripts in a chemical formula show the ratio of atoms (or ions) in a molecule a sample of CaCl 2 has 1 calcium ion : 2 chlorine ions We can use the unit “mole” to count things Atoms and ions can be counted by “moles” If the subscripts give the ratio of atoms, then they also give the ratio of moles of atoms a sample of CaCl 2 has 1 mole of calcium ions : 2 moles of chlorine ions
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Using Mole Ratio of Atoms in a Molecule Therefore, if the ratio of moles of each atom is found… then the subscripts of the chemical formula are known 1 mole C 2 mole H CH 2
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Example #3 If given percents, use those percents as grams (as if you assume you have a 100 g sample)—Remember %’s add up to 100! Change grams to moles for each atom Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles) Use the ratio as subscripts for writing the chemical formula 1 2 3 4 Example: Find the empirical formula if a sample contains Ca and Cl and is 36.1% Ca
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Example #3 If given percents, use those percents as grams (as if you assume you have a 100 g sample)—Remember %’s add up to 100! Change grams to moles for each atom Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles) Use the ratio as subscripts for writing the chemical formula 1 2 3 4 36.1 g Ca = _____ mol Ca g Ca mol Ca 1 40.08 0.901 63.9 g Cl = _____ mol Cl g Cl mol Cl 1 35.45 1.80 0.901 mol Ca = 1 mol Ca 0.901 1.80 mol Cl = 2 mol Cl 0.901 CaCl 2 Example: Find the empirical formula if a sample contains Ca and Cl and is 36.1% Ca
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Let’s Practice #3 40.92 g C = _____ mol C g C mol C 1 12.01 3.41 4.58 g H = _____ mol H g H mol H 1 1.01 4.53 3.41 mol C = 1 mol C 3.41 4.53 mol H = 1.33 mol H 3.41 C3H4O3C3H4O3 54.5 g O = _____ mol O g O mol O 1 16.00 3.41 3.41 mol O = 1 mol O 3.41 Multiply the ratio (1 : 1.33 : 1) by 3 to make a whole number ratio (3 : 4: 3) Find the empirical formula if a sample contains 40.92 g C, 4.58 g H and 54.5 g O
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Molecular Formulas
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What’s a Molecular Formula? Empirical Formula – Chemical formula determined from lab data. Lowest possible ratio of atoms Molecular Formula – Actual ratio of atoms in a molecule
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Empirical versus Molecular Formula The empirical formula is the lowest possible ratio. A molecule with the empirical formula: Could have one of the following molecular formulas: NO 2 NO 2, N 2 O 4, N 4 O 8 … CH 2 CH 2, C 2 H 4, C 4 H 8 … The molecular formula is the actual ratio
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Example #4 Find the empirical formula, if not given Find the molar mass of the empirical formula Find the ratio of the molecular formula’s molar mass (must be given to you) to the empirical formula’s molar mass Multiply the empirical formula’s subscripts by the ratio found in step 3. 1 2 3 4 Example: Empirical formula = C 3 H 4 O 3. The molecular formula’s molar mass = 176.14 g/mole. Find the molecular formula.
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Example #4 Find the empirical formula, if not given Find the molar mass of the empirical formula Find the ratio of the molecular formula’s molar mass (must be given to you) to the empirical formula’s molar mass Multiply the empirical formula’s subscripts by the ratio found in step 3. 1 2 3 4 C3H4O3C3H4O3 176.14 g/mole = 2 88.07 g/mole 3 3 12.01 g/mole 16.00 g/mole = 36.03 g/mole = 48.00 g/mole + 88.07 g/mole 41.01 g/mole = 4.04 g/mole C O H 2 C6H8O6C6H8O6 Example: Empirical formula = C 3 H 4 O 3. The molecular formula’s molar mass = 176.14 g/mole. Find the molecular formula.
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Hydrate Formulas
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What’s a Hydrate? Hydrate – Molecule that has water physically attached to it It’s not dissolved in water…hydrates can be solid, liquid or gas! BaI 2 2 H 2 O This means there are 2 water molecules attached to each BaI 2 molecules
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Hydrate & anhydride Hydrate = molecule with water molecules physically attached anhydride = molecule with water removed The water can be removed by heating the hydrate Hydrate anhydride + water heat Finding the ratio of anhydride molecules to water molecules gives you the hydrate formula
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Example #5 Find the mass of anhydride & water if not given Change mass of anhydride & water to moles Find the ratio of the moles water to moles anhydride Write the hydrate formula 1 2 3 4 Example: 2.46 g MgSO 4 hydrate is heated and 1.20 g MgSO 4 anhydride is left. Find the hydrate formula.
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Example #5 Find the mass of anhydride & water if not given Change mass of anhydride & water to moles Find the ratio of the moles water to moles anhydride Write the hydrate formula 1 2 3 4 Example: 2.46 g MgSO 4 hydrate is heated and 1.20 g MgSO 4 anhydride is left. Find the hydrate formula. 1.20 g MgSO 4 = _______ mol MgSO 4 g MgSO 4 mol MgSO 4 1 120.38 0.00997 1.26 g H 2 O = _______ mol H 2 O g H 2 O mol H 2 O 1 18.02 0.0699 0.0699 mol H 2 O = 7.01 0.00997 mole MgSO 4 MgSO 4 7 H 2 O Hydrate = anhydride + water 2.46 g = 1.20 g + water Water = 1.26 g
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What did you learn about forensic chemistry?
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Forensic Chemistry Analyze a sample Often wants to Solubility & Precipitation Solutions Chromatography Mixtures Compound Empirical & Molecular formulas
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