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<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
Describe the process by which Ag+ ions are precipitated out of solution. 4. In your testing, several precipitates are formed, and then dissolved as complexes.
