Why do we need to use amount in moles to solve stoichiometry problems? Why can't I just convert from mass to mass?
I'm not sure how to answer. I think I know the answer and just can't formulate it into words.
I have always tried to impress upon students that reactions are between molecules (particles) an when 6.02 x 10^23 particles are put together we have a mole. Reactions are between moles and not grams. Grams, of course, get into the equations because moles = g/molar mass but the reaction is between moles and not grams. We don't have balances that weigh in mols; our balances weigh in grams so ultimately we express input and output in grams but the stoichiometry is by mols. That's why we convert from grams to moles before we start the stoichiometric calculations. We CAN express the equations in grams, and by a little slight of hand, make it appear that grams are doing all the work but that is trickery. For example,
2KClO3 ==> 2KCl + 3O2
Suppose we have 12.25 grams KClO3. How much O2 will we collect? 12.25/122.5 = 0.1 mole. 0.1 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.1 x 3/2 = 0.15 moles oxygen an that is 0.15 mol x (32 g/mol) = 4.8 g oxygen. But watch what we can do.
2KClO3 ==> 2KCl + 3O2
molar mass KClO3 = 122.5 so
2 x 122.5 grams KClO3 will produce 3 x 32 grams oxygen OR 245 g KClO3 will produce 96 g oxygen. So how much oxygen will we get? We will obtain 96 grams x (12.25 g/245) = 4.8 grams. In fact, that is the way I learned to do it when I was a freshman in college MANY years ago and I used that method throughout graduate school. At first glance it appears we never make the conversion from moles to grams and that grams are going to grams. BUT, that first step of 2 x 122.5 = 245 and 3 x 32 = 96 put the mole ratios into gram ratios. From there it was just ratio and proportion. This may be more than you ever wanted to know about how molecule react but it's free information and that's the best kind.
The element nickel has ccp packing with a face-centered cubic unit cell. The volume of the unit cell is 4.38 x 10-23 cm3. Calculate the density (kg/m3) of the element.
No worries! I'm here to help you understand why using moles is crucial in solving stoichiometry problems.
Stoichiometry deals with calculating the quantities of reactants and products involved in a chemical reaction. It helps us determine the ratios of substances involved in the reaction. While it is possible to convert from mass to mass directly, using moles greatly simplifies the process and provides a more accurate approach. Here are a few reasons why moles are preferred:
1. Balanced Chemical Equations: Stoichiometry relies on balanced chemical equations, which represent the proportions of reactants and products in a reaction. These coefficients represent the number of moles involved in the reaction. By using moles, we are directly working with the ratios provided by the balanced equation.
2. Molar Ratios: The coefficients in a balanced equation can be interpreted as molar ratios. The ratio of coefficients corresponds to the ratio of moles of reactants and products. This allows us to establish a relationship between the substances involved in the reaction.
3. Conservation of Mass: Chemical reactions follow the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. By using moles, we align with this principle more effectively. Moles maintain a consistent representation of the number of particles involved, regardless of the masses of the substances.
4. Molar Mass: Converting from mass to moles requires knowledge of the molar mass of a substance. The molar mass represents the mass of one mole of a substance and is derived from the atomic masses of the elements present in the compound. By utilizing molar mass, we can relate the mass of a substance to the number of moles, making calculations easier and more precise.
In summary, using moles in stoichiometry problems provides a more efficient way to calculate the quantities of reactants and products. It allows us to work with the ratios provided by a balanced chemical equation, incorporates the law of conservation of mass, and simplifies calculations by using molar mass. So, while it is possible to convert directly from mass to mass, employing moles in stoichiometry brings greater accuracy and consistency to the calculations.