1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!
2) Mass fraction of this is excessive data.
3) The solution is:
m(Ca)=1.2 g
m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)
m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
Answer:
In thermodynamics, the triple point of a substance is the temperature and pressure at which the three phases of that substance coexist in thermodynamic equilibrium. It is that temperatureand pressure at which the sublimation curve, fusion curve and vaporisation curve meet.
In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.
Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.
To calculate molecular formula of compound, convert mass into number of moles as follows:

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.
Taking the ratio:

Putting the values,

Thus, molecular formula of compound will be
.