Chemistry Help for 2 Empirical Formulas, Please show work with answer,?

1.) Calculate the formula of a hydrated sodium carbonate given the following data.

Mass of crucible- 17.76g

Mass of Crucible and hydrate- 51.62

Mass of crucible and anhydride- 30.30

2.) Calculate the Empirical formula of Aluminum bromide given the following data.

Mass of Flask- 89.07g

Mass of Flask and Aluminum- 94.54g

Mass of Flask and Aluminum bromide- 143.14

2 Answers

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  • John
    Lv 7
    8 years ago
    Favorite Answer

    Mass of sodium carbonate hydrate = 51.62-g - 17.76-g = 33.86-g

    Mass of anhydrous sodium carbonate = 30.30-g - 17.76-g = 12.54

    Mass of water driven off in heating= 33.86-g - 12.54-g = 21.32-g H₂O

    Moles of Na₂CO₃ (106-g/mol) = 12.54-g/106-g/mol = .118-mol Na₂CO₃

    Moles of H₂O (18.0-g/mol) = 21.32-g/18.0-g/mol = 1.18-mol H₂O

    ratio of substances: Na₂CO₃ = .118/.118 = 1 ===> H₂O = 1.18/.118 = 10

    empirical formula = Na₂CO₃*10H₂O

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    Mass of Al = 94.54-g - 89.07-g = 5.47-g Al

    Mass of Br = 143.14-g - (89.07-g + 5.47-g) = 48.6-g Br

    Moles of Al = 5.47-g/27.0-g/mo = 0.203 mol Al

    Moles Br = 48.6-g/79.9-g/mol = 0.608 mol Br

    ratios: Al - .203/.203 = 1 ===> Br - .608/.203 = 2.995 ≈ 3

    empirical formula: AlBr₃

  • 4 years ago

    You have got to find the quantity of mols of every detail. To try this you use the system n=m/M (quantity of mols is the same as the mass of the element over the molecular mass of the element). Sixty three.2g of C is equivalent to about 5.Three n. 5.26g of H is equivalent to about 5.Three n. 41.6g of O is similar to about 2.6 n. For this reason, the empirical system is C2H2O.

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