Chemistry help?
2017-09-27 9:38 am
A sample of a gaseous binary compound of boron and chlorine weighing 2.842 g occupies 0.153 L. This sample is decomposed to give 0.664 g of solid boron and enough gaseous chlorine (Cl2) to occupy 0.688 L at the same temperature and pressure. Determine the molecular formula of the compound.
回答 (2)
2017-09-27 11:00 am
Relative atomic masses: B = 10.8, Cl = 35.5
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Method 1 :
Mass of B in 2.842 g of the compound = 0.664 g
Mass of Cl in 2.842 g of the compound = (2.842 - 0.664) g = 2.178 g
No. of moles of B in 2.842 g of the compound = (0.664 g) /(10.80 g/mol) = 0.0615 mol
No. of moles of Cl in 2.842 g of the compound = (2.178 g) / (35.5 g/mol) = 0.0614 mol
In the compound, mole ratio B : Cl = 0.0615 : 0.0614 ≈ 1 : 1
Hence, the empirical formula = BCl
At the same T and P:
Mole ratio (the compound : Cl₂ formed) = Volume ratio (the compound : Cl₂ formed) = 0.153 L : 0.688 L = 1 : 4.5
No. of moles of Cl₂ formed = (2.178 g) / (35.5 * 2 g/mol) = 0.0307 mol
No. of moles of 2.842 g of the compound = (0.0307 mol) * (1/4.5) = 0.00682 mol
Molar mass of the compound = (2.842 g) / (0.00682 mol) = 417 g/mol
Let (BCl)n be the molecular formula of the compound.
Molar mass of the compound in g/mol
(10.8 + 35.5) * n = 417 g
n = 9
Hence, molecular formula = B₉Cl₉
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Method 2 :
Mass of B in 2.842 g of the compound = 0.664 g
Mass of Cl in 2.842 g of the compound = (2.842 - 0.664) g = 2.178 g
At the same T and P:
Mole ratio (the compound : Cl₂ formed) = Volume ratio (the compound : Cl₂ formed) = 0.153 L : 0.688 L = 1 : 4.5
No. of moles of Cl₂ formed = (2.178 g) / (35.5 * 2 g/mol) = 0.0307 mol
No. of moles of 2.842 g of the compound = (0.0307 mol) * (1/4.5) = 0.00682 mol
Molar mass of the compound = (2.842 g) / (0.00682 mol) = 417 g/mol
In each molecule of the compound :
No. of B atoms = 417 × (0.664/2.842) / 10.8 = 9.02 ≈ 9
No. of Cl atoms = 417 × (2.178/2.842) / 35.5 = 9
Hence, molecular formula = B₉Cl₉
====
Method 1 :
Mass of B in 2.842 g of the compound = 0.664 g
Mass of Cl in 2.842 g of the compound = (2.842 - 0.664) g = 2.178 g
No. of moles of B in 2.842 g of the compound = (0.664 g) /(10.80 g/mol) = 0.0615 mol
No. of moles of Cl in 2.842 g of the compound = (2.178 g) / (35.5 g/mol) = 0.0614 mol
In the compound, mole ratio B : Cl = 0.0615 : 0.0614 ≈ 1 : 1
Hence, the empirical formula = BCl
At the same T and P:
Mole ratio (the compound : Cl₂ formed) = Volume ratio (the compound : Cl₂ formed) = 0.153 L : 0.688 L = 1 : 4.5
No. of moles of Cl₂ formed = (2.178 g) / (35.5 * 2 g/mol) = 0.0307 mol
No. of moles of 2.842 g of the compound = (0.0307 mol) * (1/4.5) = 0.00682 mol
Molar mass of the compound = (2.842 g) / (0.00682 mol) = 417 g/mol
Let (BCl)n be the molecular formula of the compound.
Molar mass of the compound in g/mol
(10.8 + 35.5) * n = 417 g
n = 9
Hence, molecular formula = B₉Cl₉
====
Method 2 :
Mass of B in 2.842 g of the compound = 0.664 g
Mass of Cl in 2.842 g of the compound = (2.842 - 0.664) g = 2.178 g
At the same T and P:
Mole ratio (the compound : Cl₂ formed) = Volume ratio (the compound : Cl₂ formed) = 0.153 L : 0.688 L = 1 : 4.5
No. of moles of Cl₂ formed = (2.178 g) / (35.5 * 2 g/mol) = 0.0307 mol
No. of moles of 2.842 g of the compound = (0.0307 mol) * (1/4.5) = 0.00682 mol
Molar mass of the compound = (2.842 g) / (0.00682 mol) = 417 g/mol
In each molecule of the compound :
No. of B atoms = 417 × (0.664/2.842) / 10.8 = 9.02 ≈ 9
No. of Cl atoms = 417 × (2.178/2.842) / 35.5 = 9
Hence, molecular formula = B₉Cl₉
2017-09-27 10:09 am
(0.664 g B) / (10.811 g B/mol) = 0.061419 mol B
((2.842 g total) - (0.664 g B)) / (35.4532 g Cl/mol) = 0.061433 mol Cl
Divide by the smaller number of moles:
(0.061419 mol B) / 0.061419 mol = 1.0000
(0.061433 mol Cl) / 0.061419 mol = 1.0002
Round to the nearest whole numbers to find the empirical formula:
BCl
((2.842 g total) - (0.664 g B)) / (35.4532 g Cl/mol) = 0.061433 mol Cl
Divide by the smaller number of moles:
(0.061419 mol B) / 0.061419 mol = 1.0000
(0.061433 mol Cl) / 0.061419 mol = 1.0002
Round to the nearest whole numbers to find the empirical formula:
BCl
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