# 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 and Aluminum- 94.54g

Mass of Flask and Aluminum bromide- 143.14

Relevance

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

• 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.