<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
There will be a shift towards the reactants
Answer:
CH4
Explanation:
The number of moles of carbon and hydrogen has been given as follows:
C = 0.300 mol
H = 1.20 mol
Next, we divide each mole value by the smallest (0.300)
C = 0.300 ÷ 0.300 = 1
H = 1.20 ÷ 0.300 = 4
The empirical ratio of Carbon and Hydrogen is 1:4, hence, the empirical formula is CH4
Answer:
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