Stoichiometric Characterization Lab Report

work backwards through stoichiometric calculations to find the mols of hydrogen peroxide in the solution. The ratio between S 2 O 3 2 – and I 3 – is 2:1, therefore there are 0.0001385 mols of I 3 – . The ratio between I 3 – and H 2 O 2 is 1:1, therefore there are 0.0001385 mols of H 2 O 2 as well. Now that we have the number of mols of hydrogen peroxide, we can convert back into grams in order to find the mass of the Na 2 CO 3 . The molar mass of H 2 O 2 is 33.96 grams, so by multiplying by that value, we find that there are 0.004703 grams of H 2 O 2 in the solution ( 33 .96 ????? ??? ???) × ( 0 .0001385 ????) = 0 .004703 ????? . Subtracting this value from the mass of the sodium percarbonate mass recorded in the very beginning of the experiment (0.027 grams) tells us that the solution contains 0.02229 grams of Na 2 CO 3 . Now, we can convert this value into mols by dividing by the molar mass of Na 2 CO 3 (82.998 grams) → (0 .02229 grams Na 2 CO 3 ) (82 .998 grams per mol ) = 0.0002686 mol Na 2 CO 3 . Now, we have the number of moles of sodium carbonate and hydrogen peroxide, we can use these to write the chemical equation: (Na 2 CO 3 ) 0.0002686 (H 2 O 2 ) 0.0001385 , we can then find the ratio between the two (n) by dividing both mol numbers by the smallest number of mols. 0 .0002686 ?𝑜? Na 2 CO 3 0 .0001385 = 1.935. This is the ratio between the two compounds. In experiment two, we want to find the mass percentage of sodium chloride in a sodium chloride + sodium nitrate solid (NaCl+ NaNO 3 ). First, we find the mass of the silver nitrate solution added to the sodium solution; this mass can be found by subtracting the mass of the solution in the Erlenmeyer flask before silver nitrate was added from the mass after the silver nitrate was added: ???? ?? 𝐴?? ? 3 ?????𝑖?? = ( ???? ????? ???𝑖?𝑖?? ?? 𝐴?? ? 3 ?????𝑖??) − (???? ?????? ???𝑖?𝑖?? ?? 𝐴?? ? 3 ?????𝑖?? ) For me, this value was 5.823 grams. Then, by multiplying this value by the percent by weight of silver nitrate in the solution (1.7% W/W), we get the mass of silver nitrate added. ( 5 .823 ????? 𝐴?? ? 3 ?????𝑖??) × 0 .017 = (0 .09899 ????? 𝐴?? ? 3 ) . Now, this value can be converted into mols by dividing by the molar mass of silver nitrate (169.374 grams): 0 .09899 𝑔?𝑎?? 𝐴𝑔? ? 3 169 .374 𝑔?𝑎?? /?𝑜? = 0.0005827 mols AgN O 3 . Through the stoichiometric ratios present in the chemical equation in which AgN O 3 reacts with NaCl, we find that the ratio of AgN O 3 to NaCl is 1:1, therefore, there are 0.0005827mols of NaCl present in the sample. Multiplying this number of mols by the molar mass of NaCl (58.44 grams), gives the mass of sodium chloride present in the sample. (0 .0005827 ??? ??𝐶? ) × (58 .44 ????? /??? ) = 0.03405 grams of NaCl. Now, by dividing this amount by the mass of sodium chloride + sodium nitrate solid, we can find the mass percentage of sodium chloride in the mixture: 0 .03405 𝑔?𝑎?? ?𝑎𝐶? 0 .05 𝑔?𝑎?? ?𝑖𝑥???𝑒 × 100% = 68 .10% ??𝐶? ?𝑦 ???? .