There's a really useful technique that can be applied to all kinds of problem solving, called Unit Analysis. In it you multiply and divide units, rather than numbers, to get the answer that has the right units for the question you are trying to answer.
In this case, the answer needs to be in units of [(number of Carbon atoms) / (given mass of Butane)].
The input information given is [(mass of Butane) x ("that we have")]
The reference information you will need is: Avogadro's number in units of [(number of molecules of anything) / (one mole of anything)], [(number of carbon atoms) / (one mole of Butane)], and the molecular mass of butane in units of [(grams mass) / (mole of butane)], derived from atomic weights.
Now, how did I know those pieces of Reference Info would be needed? That's what Unit analysis does - shows you what stuff to fit together, and how to get the answer you seek.
You started by converting the given [(mass of Butane) x ("that we have")] divided by [(mass of Butane) / (one mole of Butane)]; remember that dividing by a fraction is the same as multiplying by its inverse, so those units become: [(mass of Butane) x ("that we have")] x (moles of Butane) / (mass of Butane), and the (mass of Butane) terms cancel yielding : [(moles of Butane) x ("that we have")]. If you plug the numbers in, as you did, it comes out to: 2.50 / 58.12 = 0.0430 [(moles of Butane) x (" that we have")].
Now, how do we get from there to Number of Molecules of Butane? Multiply by Avogadro's Number, right? In Unit Analysis, this is:
[(moles of Butane) x ("that we have")] x [(number of molecules of Butane) / (one mole of Butane)] = [(number of molecules of Butane) x ("that we have")].
In numbers, that is 0.0430 x 6.023 x 10^23 = 0.2591 x 10^23 [(molecules of Butane) x ("that we have")].
Next step, how do we get from Number of Molecules of Butane that we have to Number of Carbon atoms that we have?
In Unit analysis, it is done by: [(number of molecules of Butane) x ("that we have")] x [(Number of Carbon atoms) / (one Butane molecule)] = [(Number of Carbon atoms) x ("that we have")].
In numbers, that is 0.2591 x 10^23 x 4 = 1.036 x 10^23 Carbon atoms that we have in the given mass of Butane.
Last clean-up item: precision of the answer. You were given several pieces of information, including the Reference Info you looked up, but among all those the least precise was the input mass of Butane at 2.50 grams. Your answer, then, must have 3 significant digits. I should be stated as: 1.04 x 10^23 Carbon atoms in 2.50 grams of Butane.
Note that Unit Analysis gave you two kinds of information. It pointed to how to convert from one set of units to another, thus identifying the additional information you needed to find. It also showed you how to combine those pieces (multiply, divide, add, etc) to get the units you seek. Then you plug in the numbers with the units, and it all works!
Also note this: to make the process clear I did this in three separate steps. But of course as you write this beast down on paper with units in both numerators and denominators, you can combine many multiplication and division operations into one multiple-units "fraction" and get to your answer in one step, as long as you can see all the items and understand the fit.
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