One of the ingredients in the Native American stomachache remedy derived from common chokecherry

One of the ingredients in the Native American stomachache...

Key Concepts

Combustion Analysis

Combustion analysis involves burning a compound to produce CO2 and H2O, then using the masses of these products to determine the amount of carbon and hydrogen in the original compound. This technique is fundamental in analytical chemistry for determining the elemental composition of organic compounds.

Empirical Formula Determination

The empirical formula is the simplest whole-number ratio of atoms in a compound. By converting the masses of carbon, hydrogen, and oxygen (typically derived indirectly by difference when necessary) found through combustion analysis into moles, one can deduce this ratio, providing the basic composition of the compound.

Colligative Properties

Colligative properties, such as freezing point depression, depend on the number of solute particles in a solution rather than their identity. They are useful for determining molar mass because they reflect how the solute affects the physical properties of the solvent.

Freezing Point Depression

Freezing point depression is a colligative property where the addition of a solute lowers the freezing point of a solvent. By applying the formula ?Tf = i·Kf·m, where i is the van’t Hoff factor and m is the molality, one can relate the change in freezing point to the concentration of the solute, aiding in molar mass calculations.

Molar Mass Determination

Molar mass determination using colligative properties involves calculating the number of moles of solute present based on the observed change in a physical property like the freezing point. This method is particularly useful when conventional techniques are impractical, as it links the measurable depression to the molar concentration of the solute.

Transcript

00:03 In the question we have given the mass of the organic compound that is 100 milligram and we divide 100 milligrams by thousand so we get 0 .1 gram the mass up carbon dioxide is given 220 milligrams again dividing it by 100 so the mass of carbon dioxide is 0 .2 gram where the mass of water is given as 40 .3 milligram when we divided by thousand to change it into gram we get 0 .04 of water now in the first step we will have to calculate the percentage of carbon and hydrogen and oxygen so for the determination of percentage of carbon we have formula that the mass of carbon dioxide divided by the mass of organic compound multiply by molar mass of carbon atom divided by molar mass of carbon dioxide molecule multiplied by hundred putting the value for carbon dioxide 0 .22 gram for organic compound 0 .1 gram multiplied by molar mass of carbon that is 12 gram per mole divide by the molar mass of carbon dioxide is 44 gram per mole multiply by 100 so we get the percentage that of carbon that is 60 percent similarly we can determine the percentage of how hydrogen by using the formula that mass of water divide by mass of organic compound multiply by molar mass of hydrogen divide by molar mass of water multiply by 100 putting the values so the percentage of hydrogen is equal to the mass of water that is 0 .04 gram divide by the mass of organic compound 0 .1 gram multiply by molar mass of carbon hydrogen that is 2 gram per mole while that for water it is 18 gram per mole multiply by 100 so we get the percentage of hydrogen that is 4 .44 percent now to calculate the percentage of oxygen we will subtract the percentage of carbon and hydrogen from 100 so 60 plus 4 .44 we get the percentage of oxygen that is 35 .5 percent now we have to calculate the number of moles of each element so we will divide the percentage of each element by its atomic mass so the number of moles of carbon will be equal to 60 divided by 12 that is equal to 5 moles the number of moles of hydrogen will be equal to 4 .44 divided by atomic mass that is 1 so it will be equal to 4 .44 moles similarly the number of moles of oxygen dividing its percent 35 .5 by 16 the atomic mass of oxygen so we get the percentage of oxygen that is equal to 2 .1 now to find out the simple least number ratio between the number of moles of each element you will divide the number of moles of each element by the least number of malls that is 2 .21 the number of moles of so for carbon we get the answer 2 .26 for hydrogen it is equal to 2 and for oxygen it is equal to 1 now the number of moles of carbon is not in whole number so to make it in whole number we will multiply it with a suitable number if we multiply it with 4 it will become approximately 9 so multiplying the other number of moles also by 4 for hydrogen we get the number of moles 8 and for oxygen 4 now the empirical formula of the compound empirical formula of the compound will be equal to c9 h8 this is the empirical formula of the compound now in the second part of the question we have given the weight of the salute w o2 which is equal to 0 .272 the weight of the solvent w -o -1 it is 10 cream we have also given the cryoscopic constant kf which is equal to 29 .8 degree c for molele and the depression in freezing point d delta t f which is equal to 4 .47 degree centigrade we have to calculate the molar mass of the compound so we will use the formula for the depression and freezing point delta t f is equal to kf into molality delta expanding the formula for the molality delta f will be equal to kf where molality is equal to weight of the salute divide by molecular mass of the salute multiply by thousand divide by weight of the solvent or solution...

Please give Ace some feedback