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Molar Mass Lab Report

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Conclusion: Every compound has a uniquely identifying molar mass as well as individual properties. In this lab, the purpose was to use the properties that three unknown alkali metal carbonate powers have, to identify the molar mass. For the three unknown substances, they all were white powders of about the same texture, and they all reacted when added to hydrochloric acid. Based on these properties, it is impossible to distinguish which one is which due to the similarities, so it is necessary to solve for the molar mass since that is unique to every element and compound. In this lab, 30 mL of hydrochloric acid are reacted with the limiting reactant that is the unknown powder, either A, B, or C.The amount of acid is negligible due to it not …show more content…

After the reaction finishes, the amount of CO2 that was released is calculated and then using the molemap, it is possible to solve for the amount of the unknown substance. The grams of the unknown powder are then divided by the moles of the substance to get the molar mass. This molar mass is then compared to the molar mass of the known substances. Unknown substance A is lithium carbonate since the molar mass produced in the experiment is 66.68 g/mol and the molar mass of lithium carbonate is 73.89 g/mol. Unknown substance B has a molar mass of 146.70 g/mol which is inconclusive since it is too different from the other molar masses. Unknown substance C is sodium carbonate since the molar mass found in the experiment is 110.03 g/mol which is close to the actual molar mass of sodium carbonate that is 105.99 g/mol. Substance A is lithium carbonate since the lab produced 0.33g of CO2 which can be used to calculate the moles of the unknown substance. Since the moles of the unknown are 0.007498 mol, we can divide the original mass used in the experiment (0.50g) by this to get 66.68 g/mol as the molar mass. This then is relatively close to the molar mass of lithium carbonate which is 73.89

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