a) Tartaric acid is composed of carbon, hydrogen a...

Question

a) Tartaric acid is composed of carbon, hydrogen and oxygen. A sample weighing 0.400g is subjected to combustion analysis and produced 0.469g carbon dioxide and 0.144g water. Find the empirical formula of tartaric acid.

b) The molar mass of tartaric acid is 150g/mol. Calculate its molecular formula

Jules Bruno · Accepted Answer

We’re being asked to determine the empirical formula of a compound composed of C, H, and O given the combustion analysis.Recall that in combustion analysis, a compound reacts with excess O2 to form products. For a compound composed of C, H, and O, the reaction looks like this:CxHyOz + O2 (excess) → x CO2 + y H2OThis means we need to do the following steps:Step 1: Calculate the mass of C and H in the compound.Step 2: Calculate the mass of O in the compound.Step 3: Determine the lowest whole number ratio of C, H, and O to get the empirical formula.Step 4: Get the ratio of the molar mass and empirical mass to determine the molecular formula.[readmore]Step 1: We know that the complete combustion of a 0.400g sample produces 0.469 g CO2 and 0.144 g H2O. We can do a mole-to-mole comparison to determine the mass of C and H.For C: The molar mass of CO2 is:CO2     1 C × 12.01 g/mol C = 12.01 g/mol                         2 O × 16.00 g/mol O = 32.00 g/mol                                          Sum = 44.01 g/molThere is 1 mole of C in 1 mole of CO2. Finding the mass of C:mass of C=0.469 g CO2×1 mol CO244.01 g CO2×1 mol C1 mol CO2×12.01 g C1 mol Cmass of C = 0.128 g CFor H: The molar mass of H2O is:H2O     2 H × 1.01 g/mol C = 2.02 g/mol                              1 O × 16.00 g/mol O = 16.00 g/mol                                                     Sum = 18.02 g/molThere are 2 moles of H in 1 mole of H2O. Finding the moles of H:mass of H=0.144 g H2O×1 mol H2O18.02 g H2O×1 mol H1 mol H2O×1.01 g H1 mol Hmass of H = 0.008 g HStep 2: We have 0.500 g of the sample compound. Subtracting the masses of C and H from this will give us the mass of O:mass of O = 0.400 g – (0.128 g C + 0.008 g H) = 0.264 g OStep 3: We next need to calculate the moles of each element and determine the lowest whole number ratio. The moles of C, H, and O are:moles of C=0.128 g C×1 mol C12.01 g Cmoles of C = 0.0107 mol Cmoles of H=0.008 g H×1 mol H1.01 g Hmoles of H = 0.008 mol Hmoles of O=0.264 g O×1 mol O16.00 g Omoles of O = 0.0165 mol ONow, we need to divide the number of moles of C and H by the smallest value, 0.008 mol, to get the smallest ratio of C, H, and O.0.0107 mol C0.008 mol=1 C           0.008 mol H0.008 mol=1 H         0.0165 mol O0.008 mol=2 OThe empirical formula is CHO2.Step 4: To determine the molecular formula, we need to divide the given molar mass (164 g/mol) by the mass of the empirical formula. The empirical formula mass is:CHO2      1 C × 12.01 g/mol C = 12.01 g/mol                1 H × 1.01 g/mol H = 1.01 g/mol                        2 O × 16.00 g/mol O = 32.00 g/mol                                            Sum = 45.02 g/molThe ratio is:150 g/mol45.02 g/mol=3Multiply the empirical formula by this ratio to get the molecular formula:(CHO2) × 3 = C3H3O6The molecular formula of the compound is C3H3O6

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