Sodium sulfate versus sodium sulfite, $Na_2SO_4$ versus $Na_2SO_3$. In the sulfate, sulfur expresses its maximum oxidation state, $S(+VI)$; in the sulfite, $Na_2SO_3$, we have $S(+IV)$.
Take perchlorates, and chlorates: $ClO_4^-$ versus $ClO_3^-$, i.e. $Cl(+VII)$ versus $Cl(+V)$. And then there are chlorites and hypochlorites. This naming system is a bit old-fashioned, and tends not to be used anymore.
1. No. of atoms present in second and third energy level (a) Calculate the number of $"H"$ atoms $"No. of H atoms" = 1.8 color(red)(cancel(color(black)("g H"))) × (1 color(red)(cancel(color(black)("mol H"))))/(1.008...
Step 1. Calculate the empirical formula. The empirical formula is the simplest whole-number ratio of atoms in a compound. The ratio of atoms is the same as the ratio of...
As always with these problems, it is useful to assume $100*g$ of unknown compound....and thus find an empirical formula... $"Moles of carbon"=(12.8*g)/(12.011*g*mol^-1)=1.066*mol.$ $"Moles of hydrogen"=(2.1*g)/(1.00794*g*mol^-1)=2.084*mol.$ $"Moles of bromine"=(85.1*g)/(79.90*g*mol^-1)=1.066*mol.$ And we...
To find the percentage of the metal, first determine the difference between the mass of the compound and the mass of chlorine. This will give the mass of the metal....
We can write an equation for the (complete) combustion of the substance as follows: $C_xH_yN_z + (x+y/2+z)/2O_2 to xCO_2+y/2H_2O+zNO_2$ That might look complicated, but all we're saying is that each...
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: $(30.4*g)/(14.01*g*mol^-1)$ $=$...
If we assume that the (23) of the gas has been compared to the density of $H_2$,then the molar mass of the gas becomes $46"g/mol"$ The organic compound burnt...
Your ultimate goal here is to figure out the smallest whole number ratio that exists between iron and oxygen in this unknown compound, i.e. the empirical formula of the compound....
