Figuring out the theoretical mass of reactants

Question

I'm working on my chemistry lab report. The initial instructions were to test the hypothesis that the mass of products of a reaction will equal the mass of the reactants.

Basically we where given a equation, our job was to balance and figure out the theoretical mass of reactants, and the theoretical mass of products.

There we're a total of three questions. In the first one I got the following differential values, 0.2 and 0.1. This seemed to be the most logical for me since I knew the law of conservation of mass implied that mass cannot be created nor destroyed but could be transformed. I attributed this small differential value to this transformation theory.

However, the next two questions I attempted still gave a very large differential value (40-120), which rejects my initial hypothesis.

I believe I may have missed something in the second and third question, below are my attempted solutions.

Question 1:

Initial Equation: $\ce{CH4 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{CH4 + 2O2 -> CO2 + 2H2O}$

$$ \begin{array}{ccccccc} \ce{CH4} & + & \ce{2O2} & \ce{->} & \ce{CO2} & + & \ce{2H2O}\\ 1(12+4) & + & 2(32) & \ce{->} & 1(12+32) & + & 2(2+16)\\ 1(16) & + & 2(32) & \ce{->} & 1(44) & + & 2(18)\\ 16 & + & 64 & \ce{->} & 44 & + & 36\\ \end{array} $$

Theoretical Calculation:

$$ \begin{array}{ccc} \ce{CH4}=\pu{16g} & & \ce{H2O}=\pu{36g}\\ + & \ce{->} & +\\ \ce{O2}=\pu{32g} & & \ce{CO2}=\pu{44g}\\ \end{array} $$

Actual Calculation:

$ \ce{CO2}=\pu{43.8g}\\ \ce{H2O}=\pu{35.9g}\\ $

Differential Value(Theoretical Calculation - Actual Calculation):

$ \ce{CO2}=\pu{0.2g}\\ \ce{H2O}=\pu{0.1g}\\ $

Question 2:

Initial Equation: $\ce{C2H6 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}$

$$ \begin{array}{ccccccc} \ce{2C2H6} & + & \ce{7O2} & \ce{->} & \ce{4CO2} & + & \ce{6H2O}\\ 2(24+6) & + & 7(32) & \ce{->} & 4(12+32) & + & 6(2+16)\\ 2(30) & + & 7(32) & \ce{->} & 4(44) & + & 6(18)\\ 60 & + & 224 & \ce{->} & 176 & + & 108\\ \end{array} $$

Theoretical Calculation:

$$ \begin{array}{ccc} \ce{C2H6}=\pu{60g} & & \ce{H2O}=\pu{108g}\\ + & \ce{->} & +\\ \ce{O2}=\pu{224g} & & \ce{CO2}=\pu{176g}\\ \end{array} $$

Actual Calculation:

$ \ce{CO2}=\pu{78.59g}\\ \ce{H2O}=\pu{48.2g}\\ $

Differential Value(Theoretical Calculation - Actual Calculation):

$ \ce{CO2}=\pu{97.41g}\\ \ce{H2O}=\pu{48.2g}\\ $

Question 3:

Initial Equation: $\ce{C3H8 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$

$$ \begin{array}{ccccccc} \ce{C3H8} & + & \ce{5O2} & \ce{->} & \ce{3CO2} & + & \ce{4H2O}\\ 1(36+8) & + & 5(32) & \ce{->} & 3(12+32) & + & 4(2+16)\\ 1(44) & + & 5(32) & \ce{->} & 3(44) & + & 4(18)\\ 44 & + & 160 & \ce{->} & 132 & + & 72\\ \end{array} $$

Theoretical Calculation:

$$ \begin{array}{ccc} \ce{C3H8}=\pu{44g} & & \ce{H2O}=\pu{72g}\\ + & \ce{->} & +\\ \ce{O2}=\pu{160g} & & \ce{CO2}=\pu{132g}\\ \end{array} $$

Actual Calculation:

$ \ce{CO2}=\pu{82.5g}\\ \ce{H2O}=\pu{45.0g}\\ $

Differential Value(Theoretical Calculation - Actual Calculation):

$ \ce{CO2}=\pu{49.5g}\\ \ce{H2O}=\pu{27.0g}\\ $

Not sure if it matters, but we where given this program to enter our calculations to simulate the reactions.

Answer 1

Try to answer to these questions, don't cheat, check the answer passing over the yellow box.

I think this software is confusing you always start from how much reactants you have and then calculate how much products you will have, suppose you have $\pu{100g}$ of oxygen and $\pu{44g}$ of propane.

How many mole of propane and oxygen ($\ce{O2}$) do you have?

Oxygen moles$=\frac{\pu{100g}}{\pu{32g/mol}}\sim \pu{3mol}$, Propane moles$=\frac{\pu{44g}}{\pu{44.096g/mol}}\sim \pu{1 mol}$

If this is your balanced equation: $$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$ How many oxygen molecules do you need to react with one molecule of propane?

You need 5 oxygen molecules for every propane molecule. So you need 5 moles of oxygen for every mole of propane.

Which is the ratio?

$\frac{1}{5}$ if you have $x$ moles of oxygen you need $\frac{x}{5}$ moles of propane. e.g. if you have 5 moles of oxygen you need? 1 mole of propane.

But you have only 3 moles of oxygen to get the products, oxygen is a limiting reagent! So how many moles of propane will react:

$$\pu{\frac{3}{5}mol} =\pu{0.6 mol} $$

So how many moles of the products will you have?

$3\times0.6$ moles of $\ce{CO2}$ and $4\times0.6$ moles of water

And so how many grams?

for $\ce{CO2}$, $\pu{1.8mol}\times \pu{44g/mol}$ for water, $\pu{2.4mol}\times \pu{18g/mol}$

Answer 2

$\pu{0.1 gram}$ is ($\pu{E4 kg}$)($\pu{299792458 m/s2}$) = $\pu{8.988E12 J}$ or two kilotons nuclear yield equivalent. Let's be conservative and say "experimental error."

Mass-energy is conserved because time is homogeneous throughout the universe. Run this continuous symmetry through Noether's theorems to obtain the First Law of Thermodynamics. However, time is not locally homogeneous! We live at the bottom of a gravitational potential well. GPS atomic clocks run slower in orbit for special relativity (orbital velocity), but faster still for general relativity (higher in the potential well). Therefore, don't transform "missing mass" except in physics labs. Their party, their mess.