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Why do you have to convert the reactants and products of an equation to moles when doing stoichiometry problems? Why can't you just convert straight to grams to solve the problem?

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Equations tell you how much of each molecule you need in relation to others. If you used grams the numbers would change because not every molecule weighs the same.

E.g. if you have $200$ hydrogen and $100$ oxygen molecules you could make $100$ water molecules. But if you had $200\ \mathrm{g}$ hydrogen and $100\ \mathrm{g}$ oxygen, you don't know how many grams of water that gives. You need to convert using the weight.

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Grams depends on the molecule itself, oxygen weighs more than hydrogen. But using moles gets rid of the mass concern, because a mole represents a specific amount of molecules, independant of weight. Since you want to convert from reactants to products, you need to know how many molecules react with other molecules, not mass. Mass doesn’t affect the outcome of a reaction, atom counts do.

Looking at a balanced equation, you’re looking at atoms. For example, $\ce{H2 + O -> H2O}$ (incorrect equation but for demonstration purposes) means that for every 2 atoms of hydrogen, and every 1 atom of oxygen, you get one molecule of $\ce{H2O}$. Now because the atoms weight different amounts, this is not the same as saying 2 grams of hydrogen and 1 gram of oxygen gives 1 gram of water. First of all, you can’t react 3 grams and end up with 1 gram, and also if you do the stoich, you have far more of one atom that the other.

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You have to use moles as the coefficients in a chemical equation refer to the molar quantity. If the coefficients were given in units of mass instead, you could use mass ratios. However, this is not the convention of chemical equations. Thus, you must use molar ratios.

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The answer is because in chemical reactions there is a conservation of moles, rather than mass. Moles are the total amount of molecules in the system, those remain constant. Naturally using numbers like $6.022 \times 10^{23}$ is inconvenient, so moles help with keeping track. Mass on the other hand, which is equivalent to energy in the system, is not conserved in systems that are not isolated (as most in real life are not).

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  • $\begingroup$ Not quite right. Overall mass is conserved in chemical reactions. In fact moles are not conserved. $\endgroup$ – MaxW Oct 11 '16 at 3:22
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jiggunjer is right that chemical equations tell you how much of each molecule you need in relation to others. "How much" in the sense of an integer count.

So consider the simple reaction $$\ce{2H2 + O2 -> 2H2O}\tag{1}$$ Thus two molecules of hydrogen and one molecule of oxygen yields two molecules of water. Now we can create any integer multiple of that reaction. So the following are also true. $$\ce{200H2 + 100O2 -> 200H2O}\tag{2}$$ $$\ce{2\times10^6*H2 + 1\times10^6O2 -> 2\times10^6H2O}\tag{3}$$ $$\ce{2billion H2 + 1 billion O2 -> 2 billion H2O}\tag{4}$$

So in the first equation above the multiple is 100, the second uses $1 \times 10^6$, and the third one billion.

The problem with atoms or molecules is that they have an exceedingly small mass. So for things on an atomic scale it is convenient to count in terms of moles. One mole of atoms or molecules, which is $6.022 \times 10^{23}$ of them, gives a mass large enough for us humans to weigh and manipulate. So if we let $\ce{N_A}$ be equal to Avogadro's number, then we can say that $$\ce{2N_AH2 + N_AO2 -> 2N_AH2O}\tag{5}$$ or in plain English two moles of hydrogen and one mole of oxygen react to produce two moles of water.

Now lets use equation 1 and multiply each side by "1 gram(s)" as we can do in algebra. We get: $$\ce{2~grams~H2 + 1~gram~O2 -> 2~grams~H2O}\tag{6}$$ In chemistry mass is conserved. This equation has a very, very large problem. There are three grams of matter on the left but only two grams on the right. The missing gram of matter isn't the energy in a balloon pop, but rather the energy of a very sizeable nuclear bomb.

Now we could try writing the equation in terms of the mass of the reactants and products, which is sometimes done. $$\ce{4~grams~H2 + 32~grams~O2 -> 36~grams~H2O}\tag{7}$$ Now if we try to use our algebra trick again we can divide both sides by "4 grams" to get: $$\ce{H2 + 8O2 -> 9H2O}\tag{8}$$ but we can see that the stoichiometry is wrong for this equation.

Now let's digress in another direction. From Wikipedia "The gram, 1/1000 of a kilogram, was provisionally defined in 1795 as the mass of one cubic centimeter of water at the melting point of ice. The final kilogram, manufactured as a prototype in 1799 and from which the International Prototype of the Kilogram (IPK) was derived in 1875, had a mass equal to the mass of 1 dm3 of water at its maximum density, approximately 4 °C." So the notion of a kilogram was created by humans as some arbitrary mass long before the notion of an atom or molecule was truly understood by scientists. Thus chemistry was tied into the existing system of measures.

Just as a mass of 1 gram is arbitrary, so too is Avogadro's number arbitrary. It has been defined now for a long time as the number of atoms of carbon in 12 grams of $\ce{^{12}C}$. But some other element/isotope could have been chosen. So 12 grams of $\ce{^{12}C}$ and 16 grams of $\ce{^{16}O}$ do not have the same number of atoms because of the binding energy in the nucleus.

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You could use mass ratios if you wanted but the convention among chemists is to use molar ratios since they are easier to work with.

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