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Chemistry Final Project

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  • Question #6 on the Fall Term Exam
  • By Margaret Akey
  • The Question States:
  • 0.46 g of an unknown organic compound containing C, H, and N is burned and produces 1.113 g of carbon dioxide and 0.34 g of water
  • Total Mass of Organic Compound: 0.46 g The reaction produces 1.113 g of CO/2 The reaction produces 0.34 g of H/2O
  • So what do we know?
  • We also know that the organic compound, containing C, H, and N is BURNED. And the problem tells us that one of the products of this reaction is CO/2 and another is H/2O
  • Since the word "burned" signifies a combustion reaction, we know that the organic compound is added with O/2 to make CO/2, H/2O and since we have Nitrogen, it also makes NO. So now we can write the reaction as: C/xH/yN/x + O/2 => CO/2 + H/2O + NO
  • C/xH/yN/x + O/2 => CO/2 + H/2O + NO
  • We write our organic compound like this because we do not know the ratio of Carbons to Hydrogens to Nitrogens (Yet!) So we represent the numbers of each with variables.
  • This leads us to our first question that we have to answer: What is the empirical formula for the compound?
  • First we have to figure out what the "empirical formula" of a compound really is. The Empirical Formula of a compound is defined as:
  • A formula that gives the simplest whole-number ratio of atoms in a compound.
  • Steps to solve an empirical formula problem: 1) Find moles of each element in your compound. 2) Calculate the (whole number) mole ratios between the elements by dividing the two bigger masses by the smallest mass. 3) Use these whole # mole ratios to generate your empirical formula by plugging them into your variables (i.e x, y and z in C/xH/yN/z)
  • Law of Conservation of Mass: Mass is neither created nor destroyed in chemical reactions.
  • 1.113 g CO/2
  • First we should find the moles of Carbon in our C/xH/yN/z .
  • This is our grams of CO/2 that we produced. Using our reaction, ( C/xH/yN/x + O/2 => CO/2 + H/2O + NO ) , we can see that carbon only appears in one of the reactants and one of the products! Therefore, we can use the law of conservation of mass to determine the moles of carbon that are in our unknown organic compound.
  • The law of conservation of mass allows us to set up this factor label, which lets us find moles of carbon, in the product, which we know is equal to the moles of carbon in the reactant.
  • 1.113 g CO/2
  • Stated in an easier way, the conservation of mass states that "the the mass of any one element at the beginning of a reaction will equal the mass of that element at the end of the reaction."
  • 1 mol CO/2
  • 44.011 g CO/2
  • 1 mol C
  • 1 mol CO/2
  • =
  • This conversion factor comes from adding the grams of one mole of carbon, 12.011 g, with the grams of two moles of oxygens, 16.00 g + 16.00 g , to get the total grams of one mol of a CO/2 molecule, 44.011 g CO/2
  • This portion of our factor label comes from the fact that there is one Carbon, in one CO/2 molecule Hence the ratio: one mol C / 1 mol CO/2
  • 0.02529 mol Carbon
  • Great! So we have found our moles of Carbon in our reaction. Now lets find moles of Hydrogen!
  • C/xH/yN/x + O/2 => CO/2 + H/2O + NO
  • Just like the carbon, hydrogen is only in one of the reactants and one of the products, so we know that ALL of the moles of hydrogen that are in the 0.34 g of H/2O are the moles of hydrogen that are in the C/xH/yN/z
  • the moles of hydrogen here...
  • equal the moles of hydrogen here according to the conservation of mass
  • 0.34 g H/2O
  • Our reaction produces 0.34 g of H/2O
  • 1 mol H/2O
  • 18.0158 g H/2O
  • This conversion factor comes from the fact that in 1 mol of water, there are two hydrogens, which each have 1.0079 g per mole, and one oxygen, which has 16.00 g per mole. So, 1.0079 + 1.0079 + 16.00 = 18.0158 g H/2O per 1 mol of H/2O
  • 2 mol H
  • 1 mol H/2O
  • =
  • This conversion factor comes from the fact that in one H/2O molecule, there are two hydrogens, hence the ratio, 2 mol H in every 1 mol of H/2O
  • 0.038 mol H
  • So now we have: moles of C: 0.02529 and moles of H: 0.038
  • Next, we need to find moles of nitrogen!
  • Remember: mass of our organic compound (C/xH/yN/x) = 0.46 g
  • We already have the moles of Carbon, and the moles of Hydrogen in our organic compound. SO... if we find the grams of carbon and hydrogen in our sample, we can subtract that from the total mass to get the mass of the nitrogen! And from there we can convert the grams of nitrogen to moles of nitrogen.
  • 0.02529 mol C
  • Convert moles of Carbon to grams of Carbon:
  • 0.3037 g C
  • 1 mol C
  • 12.011 g C
  • =
  • 0.038 mol H
  • Convert moles of Hydrogen to grams of Hydrogen
  • 0.038 g H
  • 1.0079 g H
  • 1 mol H
  • =
  • 0.46 g - (0.3037 g C + 0.038 g H)
  • 0.1182 g N
  • Now that we have found the grams of carbon and thr grams of hydrogen we can add them together and subtract them from the total mass of the unknown organic compound to get grams of nitrogen!
  • =
  • 0.1182 g N
  • 0.008442 mol N
  • Now we can convert the grams of nitrogen to moles of nitrogen.
  • 1 mol N
  • 14.007 g N
  • =
  • Now we have: moles of C: 0.02529 moles of H: 0.038 and moles of N: 0.008442
  • Knowing this information we can calculate the mole ratios between C, H and N to find the empirical formula!
  • mol C
  • mol N
  • Keep in Mind: moles of C: 0.02529 moles of H: 0.038 and moles of N: 0.008442
  • =
  • 0.02529 mol C
  • 0.008442 mol N
  • =
  • plug into calculator!!
  • 6
  • 2
  • mol H
  • Keep in Mind: moles of C: 0.02529 moles of H: 0.038 and moles of N: 0.008442
  • mol N
  • =
  • 0.038 mol H
  • 0.008442 mol N
  • =
  • plug into calculator!
  • 9
  • 2
  • Now that we have figured out that the whole # mole ratio, C : H : N is 6 : 9 : 2 we know that the empirical formula for this unknown organic compound is: C/6H/9N/2
  • YAY! We solved the first part of this problem! Next, we have some more information, and another part to solve.
  • If the root mean square velocity of the gas is 2.612 times slower than oxygen gas, what is the molecular mass of the compound?
  • Information: The root mean square velocity of the gas (the compound) is 2.612 times SLOWER than oxygen gas. What we need to find: molecular mass of the compound
  • Molecular Mass: the sum of the atomic masses in a molecule.
  • First we should go over some definitions.
  • Root Mean Square Velocity: the measure of the velocity of particles in a gas
  • Now you might think, well we have the empirical formula! To find the molecular mass we should just add up all of the masses of the atoms in C/6H/9O/2. However, that is how you would find the EMPIRICAL MASS. To find molecular mass with our given data, we have to use a formula.
  • THE FORMULA:
  • μ of oxygen
  • μ of the compound
  • where μ = the root mean square velocity and MM = molecular mass
  • =
  • √
  • (
  • MM of compound
  • MM of Oxygen
  • )
  • μ of the compound
  • μ of oxygen
  • The x's will cancel! :)
  • =
  • 2.612 x
  • x
  • SInce the compound's root mean square velocity is 2.612 times SLOWER than that of oxygen gas... we know that oxygen gas is 2.612 times FASTER than the compound! Therefore we can make oxygen = 2.612 * x and the compound = x
  • Since the x's will cancel, we will end up with
  • 2.612
  • since it is oxygen gas, the molecular formula is O/2 so when we calculate the molecular mass of O/2 we have to account for 2 oxygens. 16.00 + 16.00
  • =
  • √
  • (
  • MM of compound
  • 32.00
  • )
  • 2.612
  • 6.823
  • To solve for the MM of the compound: 1) square both sides
  • =
  • =
  • √
  • MM of compound
  • (
  • 32.00
  • MM of compound
  • 32.00
  • )
  • Then, we multiply both sides by 32.00 SO... the MM of the compound = 218.32
  • And we have to watch sig digs, so since we have 4 sig digs that makes our answer...
  • Molecular Mass of the unknown compound =
  • 218.3 a.m.u.
  • Now on to the last part of this problem!! Almost done! :) Next, we have to find the Molecular Formula of the compound.
  • Difference between molecular and empirical formula: The molecular formula is an expression of the number and type of atoms that are present in a single molecule of a substance. The empirical formula is the simplest ratio of elements present in the compound.
  • To find the molecular formula, 1) find the ratio of molecular mass to empirical mass by dividing MM by EM 2) Multiply each of the subscript #'s in the Empirical Formula by the ratio you found in the previous step! (This is the Molecular Formula)
  • Molecular Mass
  • Empirical Mass
  • now multiply each of the subscripts in C/6H/9N/2 by 2 to get our molecular formula:
  • =
  • C/12H/18N/4
  • 1
  • 2
  • =
  • 2
  • I hope this was very clear and you learned a lot because this took me 6 hours to make.
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