One mole of Mg$Cl_2$ , gives one mole of Magnesium , and one mole of Chlorine gas. In terms of mass , 95 g of Mg$Cl_2$ on decomposition gives 24 g of Magnesium, 71 g of Chlorine.
Mass percentage of Mg = (mass of Mg / mass of Mg$Cl_2$ ) x 100
mass percentage of Mg = ( 24 g / 95 g) x 100 = 25.26 %
Mass percentage of Cl = (mass of Cl / mass of Mg$Cl_2$ ) x 100
mass percentage of Cl = ( 71 g / 95 g) x 100 = 74.74 %
An empirical formula represents the lowest whole number ratio of in a compound. Since the percentages of the elements equal 100%, we can assume we have a 100 g sample,...
In 100 g of compound there are 47.0 g potassium, 14.5 g carbon, and 38.5 g oxygen. Why? We divide thru by the atomic masses: $K$ $=$ $(47.0*g)/(39.1*g*mol^(-1))$ $=$ $1.20$....
As is typical with these questions, we assume $100*g$ of unknown compound, and work out the MOLAR quantities of each element present: $" moles of C":$ $(15.8*g)/(12.011*g*mol^-1)=1.32*mol$ $" moles of...
In $100* g$ of this there are $30.4* g$ $N$, and the balance $O$. We divide thru by the atomic masses in order to approach the empirical formula: $(30.4*g)/(14.01*g*mol^-1)$ $=$...
The of tin in the compound will be equal to the ratio between the mass of tin and the mass of the compound, multiplied by $100$. $color(blue)("% Sn"...
And then we work out the molar composition. Thus $"Moles of carbon"$ $=$ $(92.3*g)/(12.011*g*mol^-1)=7.68*mol$. And $"Moles of hydrogen"$ $=$ $(7.7*g)/(1.00794*g*mol^-1)=7.64*mol$. Clearly, there is $1:1$ molar equivalence with respect to $C:H$....
As with all these problems, we assume a $100*g$ mass of unknown compound, and then we work out the molar quantity: $"Moles of potassium"=(47.0*g)/(39.10*g*mol^-1)=1.20*mol$ $"Moles of carbon"=(14.5*g)/(12.011*g*mol^-1)=1.21*mol$ $"Moles of oxygen"=(38.5*g)/(16.0*g*mol^-1)=2.41*mol$...