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
- JohnLv 78 years agoFavorite 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
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
- katsaounisLv 44 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.