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c12h24
Molar Mass
The mass of 35 gmoles of carbon monoxide ( CO ) is given by : m = ( n ) ( M ) m = ( 35 gmol ) ( 28.00 g / gmol ) = 980 g <----------------
mass of h = 8 atoms of h/molecule * 1.00794 g/mol = 8.06352g/mol of h in c7h8 mass of c7h8 = 8 atoms of H/molecule * 1.00794 g/mol + 7 atoms/molecule of C * 12.0107 g/mol = 92.13842 g/mol h / c7h8 = 8.06352 / 92.13842 = 0.08752 0.08752 = 8.752% = 9%
86.62(8)%The molar masses of the constituent elements must be known.M(Pb) = 2.072(1) x 102 gmol-1M(O) = 1.59994(3) x 10 gmol-1From these and their respective molar ratios in the compound, the total molar mass of the compound must be calculated.M(PbO2) = M(Pb) + 2M(O)M(PbO2) = (2.072 x 10 gmol-1 + 2(1.59994 gmol-1)) x 10M(PbO2) = 2.392(1) x 102 gmol-1From the molar mass of the element in question and its ratio, and the compound, their total masses present in one mole of compound must be calculated.m = nMm(Pb) = 1 mol x 2.072 x 102 gmol-1m(Pb) = 2.072(1) x 102 gm(PbO2) = 1 mol x 2.392(1) x 102 g (always 1 mole)m(PbO2) = 2.392(1) x 102 gFrom these, the percentage of the element in question present in the compound can be calculated.m(Pb) x 100%/m(PbO2) = 2.072 g x 100%/2.392 gm(Pb) x 100%/m(PbO2) = 86.62(8)%
C5h10
C3H6
c12h24
c3h6
c12h24
This formula is for magnesium chloride hexahydrate: MgCl2.6H2O.
Cs2O2
The empirical formula of SN has a formula unit mass of the sum of the gram atomic masses of nitrogen and sulfur, i.e., about 46.0667. The gram molecular mass given in the problem divided by this formula unit mass is about 4. Therefore, the molecular formula is S4N4.
Cs2O2
The average molecular weight of dry air is 28.96 g/gmol.
i got 23765.4
Given: M.M. Carbon - 12.0107; M.M. Hydrogen - 1.00794 Solution: The molecular mass of C7H8 is = (6 x 12.0107) + (7 x 1.00794) = 79.11978 g The total mass of hydrogen present is = (7 x 1.00794) = 7.05558 g Therefore, the % m/m of hydrogen is = (7.05558 g / 79.11978 g) x 100% = 8.9%