## Double Check my Understanding

Brendan Duong 1I
Posts: 69
Joined: Wed Sep 30, 2020 10:07 pm

### Double Check my Understanding

Hi, so just to double check my understanding, when calculating molar mass of compounds, we ignore the coefficient in front of the compound (ie. if there is a two, we don't multiply molar mass by two) because the coefficient simply represents the molar ratio of reactants/products, at least when we are doing limiting reactant calculations. If so, is there anytime we do multiply that coefficient to the molar mass when trying to solve anything?

JoshMoore2B
Posts: 60
Joined: Wed Sep 30, 2020 9:51 pm

### Re: Double Check my Understanding

You are indeed correct: when calculating the molar mass of a compound, you do not take into account the stoichiometric coefficient in front of the compound in the balanced chemical equation.

A situation you would use the stoichiometric coefficient times the molar mass of a compound would be when you need to find the limiting reactant in an equation. Suppose you had two compounds A and B, both reactants, and you need 2 moles of A for the reaction to occur and 3 moles of B. Also suppose that the question gave you the amount of A and B, in moles, that you used to produce a reaction. To find the limiting reactant, you would find 2 * molar mass of A and 3 * molar mass of B, and then divide the stoichiometric coefficients (in this case 2 and 3) by the amount of A and B you had in moles, Whichever is greater (i.e. A divided by the amount you had is greater than B divided by the amount you have) is the compound in excess, while the other is the limiting reactant.

Hope this helps!

Sabrina Galvan 3A
Posts: 64
Joined: Wed Sep 30, 2020 9:35 pm
Been upvoted: 1 time

### Re: Double Check my Understanding

Brendan Duong 1I wrote:Hi, so just to double check my understanding, when calculating molar mass of compounds, we ignore the coefficient in front of the compound (ie. if there is a two, we don't multiply molar mass by two) because the coefficient simply represents the molar ratio of reactants/products, at least when we are doing limiting reactant calculations. If so, is there anytime we do multiply that coefficient to the molar mass when trying to solve anything?

You are correct! When only calculating the molar mass, we do not consider the coefficients because they would give us a value that is not the true equivalent of one mole, instead it would be a value about the mole of the compound. The only numbers we consider are the subscripts behind each singular element, which gives further information on not only a compound's molecular formation, but also the weight of that compound (i.e. if a compound contains N subscript 2, then we multiple the mass of N by two, 14.007*2=28.014). Note that the two plays a role as a subscript, not a coefficient. For the limiting reactant problems we have seen so far, we use the coefficient as an intermediate step in finding the limiting reactant, because as you said, it is a representation of the molar ratio between reactants and products. Therefore it is used to solve a problem, but it is likely not your last step in solving it.

Chance Herbert 3A
Posts: 61
Joined: Wed Sep 30, 2020 9:55 pm

### Re: Double Check my Understanding

You are correct! When you calculate the molar mass of an element or molecule you will not multiply this by the stoichiometric coefficient; however, stoichiometric coefficients are relevant in determining the number of moles of each reactant and product present in a reaction. Take for example the reaction N2 + 3H2 ->2NH3. If you were to determine that Nitrogen gas were the limiting reactant and were given or calculated the moles of this reactant, you could find the moles of both H2 and NH3 present in the reaction. You would multiply the number of moles of N2 by 3 to find the number of moles of H2 consumed and by 2 to find the number of moles of NH3 formed. Hope this helps :)

Kaihan_Danesh_2J
Posts: 64
Joined: Wed Sep 30, 2020 9:40 pm

### Re: Double Check my Understanding

You are right. To put it simply, the stoichiometric coefficients are there to create balance in the chemical reaction, not to denote the molar mass of the compounds. So, if you were to find the molar mass of a compound, you would just add the up the masses of the elements without multiplying them by the stoichiometric coefficient.

Inderpal Singh 2L
Posts: 107
Joined: Wed Sep 30, 2020 10:02 pm
Been upvoted: 2 times

### Re: Double Check my Understanding

Yup, you are good. The stoichiometric coefficients are just there to create a balance between the two sides and don't actually go into the calculation.

Ayesha Aslam-Mir 3C
Posts: 83
Joined: Wed Sep 30, 2020 9:43 pm

### Re: Double Check my Understanding

Hello, I also would like to pose a question in terms of understanding with limiting reactant questions; if we ignore the stoich coefficients, we would be focusing on just the mass, correct? So while the moles of a compound do indicate the amount of an element in terms of mass, we still need to keep track of moles in ensuring a chemical equation is balanced. So if you are given a certain mass of some reactant and, the same mass of the product would be created, which means the same moles are on both sides of the equation?

Edward Castro
Posts: 50
Joined: Wed Sep 30, 2020 9:32 pm

### Re: Double Check my Understanding

You are right on track! By adding. the coefficients you are creating a balance that allows you to get closer towards finding the result that you want.

David Y
Posts: 57
Joined: Wed Sep 30, 2020 9:49 pm

### Re: Double Check my Understanding

You are correct, you do not want to multiply the molar masses by the coefficients.

Nan_Guan_1L
Posts: 64
Joined: Wed Sep 30, 2020 9:59 pm

### Re: Double Check my Understanding

Ayesha Aslam-Mir 2E wrote:Hello, I also would like to pose a question in terms of understanding with limiting reactant questions; if we ignore the stoich coefficients, we would be focusing on just the mass, correct? So while the moles of a compound do indicate the amount of an element in terms of mass, we still need to keep track of moles in ensuring a chemical equation is balanced. So if you are given a certain mass of some reactant and, the same mass of the product would be created, which means the same moles are on both sides of the equation?

it doesn't necessarily have to be the same moles on both side of the equation. Yes the mass of the reactant and product are the same, but they usually are different substances and have different molar mass, which means that their mass per mol is different. this would create different moles on both side of the equation although their mass in total is the same. I hope this answers your question!

Ellison Gonzales 1H
Posts: 64
Joined: Wed Sep 30, 2020 10:00 pm

### Re: Double Check my Understanding

JoshMoore3D wrote:You are indeed correct: when calculating the molar mass of a compound, you do not take into account the stoichiometric coefficient in front of the compound in the balanced chemical equation.

A situation you would use the stoichiometric coefficient times the molar mass of a compound would be when you need to find the limiting reactant in an equation. Suppose you had two compounds A and B, both reactants, and you need 2 moles of A for the reaction to occur and 3 moles of B. Also suppose that the question gave you the amount of A and B, in moles, that you used to produce a reaction. To find the limiting reactant, you would find 2 * molar mass of A and 3 * molar mass of B, and then divide the stoichiometric coefficients (in this case 2 and 3) by the amount of A and B you had in moles, Whichever is greater (i.e. A divided by the amount you had is greater than B divided by the amount you have) is the compound in excess, while the other is the limiting reactant.

Hope this helps!

This explanation helped, thank you! For your specific example of A and B, do you divide A by 2 or 3? I confuse myself with this topic on which coefficient to divide by. Thank you!