<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol.
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.
C H O
Mass in 100 g 40.0 g 6.7 g 53.5 g
Molar mass 12 g/mol 1 g/mol 16 g/mol
Number of moles 40.0/12= 3.33 6.7/1 = 6.7 53.5/16 = 3.34
Divide by the least number of moles
3.33/3.33 = 1 6.7/3.33 = 2.01 3.34/3.33 = 1.00
after rounding off
C - 1
H - 2
O - 1
Empirical formula - CH₂O
Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O)
Molecular formula - C₂H₄O₂
Answer:
32.7 kilograms of aluminium oxide will be produced.
Explanation:
Mass of aluminum = 17.3 kg = 17300 g (1 kg = 1000 g )
Moles of aluminium = 
According to reaction, 2 moles of aluminum gives 1 mole of aluminum oxide,then 640.74 moles of aluminum will give:
of aluminum oxide
Mass of 320.37 moles of aluminum oxides:
320.37 mol × 102 g/mol = 32,677.74 g = 32.67774 kg ≈ 32.7 kg
32.7 kilograms of aluminium oxide will be produced.
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Copper (I) oxidation state is 1 Cu2So4
copper (II) oxidation state is +2 CuSo4
copper (i) also give up one electron so you need two of them to react with the sulfate ion (which has charge of -2)
and also all metallic ions have an multiple oxidation levels corresponding to the number of electrons they can exchange or loose
Hope this helps
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