Presentation on theme: "Calculating Empirical Formulas"— Presentation transcript:
1Calculating Empirical Formulas Lowest whole-number ratio of the atoms of the elements in a compoundC6H12O6 (glucose)The ratio that glucose normally has for carbon:hydrogen:oxygen is 6:12:6.The lowest ratio that glucose has for carbon:hydrogen:oxygen is 1:2:1 (each number can be divided by the smallest number in the ratio which is 6).
2Next - Calculating empirical formulas The empirical formula for glucose is CH2O since this is the lowest whole-number ratio of atoms for that compound.May or may not be the same as the normal molecular formula of a compoundNext - Calculating empirical formulas
3What is the empirical formula of a compound that is 25 What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?If 25.9% of the compound is nitrogen and 74.1% of the compound is oxygen, then a compound with a mass of 100 g has 25.9 g of nitrogen and 74.1 g of oxygen.To calculate the empirical formula, we need to relate the moles of each atom in the compound, so we need to convert the masses of the elements to moles.
4This would mean that the ratio of nitrogen to oxygen is N1.85O4.63.We can divide each number in the ratio of N1.85O4.63 by1.85 to get N1O2.50.Since we cannot have 2.50 atoms ofoxygen, we must multiply through each number by 2 toeven it out, getting N2O5 as our empirical formula.
5In calculating empirical formulas, remember that the number of atoms is a whole number. If the number of atoms for an element is close to a whole number (i.e., 2.1 or 2.2 or 2.8, or 2.9), you can usually round up or down to get a whole number.If you should get a number of atoms closer to 2.33 or 2.5, multiply each number in the formula by a number that gets that to a whole number. For example, if you calculated 2.33, you would multiply this by 3 to get a value of 7 for that number.
6Give it a tryDetermine the empirical formula for a compound containing 7.8% carbon and 92.2% chlorine.7.8 g C 1 mol C12.0 g C = 0.65 mol C92.2 g Cl 1mol Cl35.5 g Cl = mol ClC0.65Cl2.60 divide each by 0.65Final answer = CCl4
8Calculating Molecular Formulas Although sometimes a molecular formula may be the same as a molecule’s empirical formula, like in carbon dioxide (CO2), we have seen that the empirical formula for glucose is not the same as its molecular formula.One can determine the molecular formula of a compound by knowing its empirical formula and its mass.Next - Example
9Calculate the molecular formula of the compound whose molar mass is g and empirical formula is CH2O.We know that the molecular formula will have a molarmass of g. We also know, by calculating thegmm of CH2O, that CH2O has an empirical formulamass (efm) = g CH2O.Now, in order to figure out what we must multiply eachnumber in the empirical formula by, we must figure out bywhat number we must multiply the empirical formula massto get the molecular formula mass.To get from to ,
10Therefore, we must multiply each number of atoms in CH2O by 6 to get the molecular formula of C6H12O6.You can double-check your answer by recalculating the molar mass of C6H12O6.gmm C6H12O6 = 6 x g + 12 x g + 6 x g = g C6H12O6This agrees with the molar mass we were given, so the molecular formula we calculated is correct.
11Give it a tryDetermine the molecular formula of a compound that is 40.0% C, 6.6% H, and 53.4% O and the molar mass is 120.0g.mol C = mol H = 6.6 mol O = 3.34divide each by 3.33 C1H2O1Empirical Formula weight is 30.0Take divided by 30.0 = 4Than multiply your subscripts.Final Answer is C4H8O4
12Example (toughy)1.00 g of menthol on combustion yields g of H2O and g of CO2. What is the empirical formula?Solution:1.161 g H2O g H = mol H2.818 g CO2 g C = mol Cdifference = g O = mol Oratios: H/O = 20.4 and C/O =10.1Therefore, C10H20O