In relation to your example, dichloromethane has 2 protons per molecule so first you have to divide the integral by 2. then mol% = 0.01 / (1.0 + 0.01) = 0.99mol% ie you divide by the sum of all the components. This doesn't answer your question fully but I hope it is helpful.helpful. To calculate wt% you must sum the products of the integral and MW ie, 0.01MW(DCM) / (0.01MW(DCM) + 1.0*MW(Compound1) + ......)

In relation to your example, dichloromethane has 2 protons per molecule so first you have to divide the integral by 2. then mol% = 0.01 / (1.0 + 0.01) = 0.99mol% ie you divide by the sum of all the components. This doesn't answer your question fully but I hope it is helpful. To calculate wt% you must sum the products of the integral and MW ie, 0.01MW(DCM) / (0.01MW(DCM) + 1.0*MW(Compound1) + ......)

In relation to your example, dichloromethane has 2 protons per molecule so first you have to divide the integral by 2. then mol% = 0.01 / (1.0 + 0.01) = 0.99mol% ie you divide by the sum of all the components. This doesn't answer your question fully but I hope it is helpful. To calculate wt% you must sum the products of the integral and MW ie, 0.010.01 * MW(DCM) / (0.01(0.01 * MW(DCM) + 1.0*MW(Compound1) + ......)