# 0.1 Relative atomic masses and empirical formulae  (Page 5/5)

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Therefore, the atomic masses of non-gaseous elements can be determined by mass and volume measurements ongaseous compounds containing these elements. This procedure is fairly general, and most atomic masses can be determined in thisway.

## Moles, molecular formulae and stoichiometric calculations

We began with a circular dilemma: we could determine molecular formulae provided that we knew atomic masses,but that we could only determine atomic masses from a knowledge of molecular formulae. Since we now have a method for determining allatomic masses, we have resolved this dilemma and we can determine the molecular formula for any compound for which we have percentcomposition by mass.

As a simple example, we consider a compound which is found to be 40.0% carbon, 53.3% oxygen, and 6.7% hydrogenby mass. Recall from the Law of Definite Proportions that these mass ratios areindependent of the sample, so we can take any convenient sample to do our analysis. Assuming that we have 100.0g of the compound, wemust have 40.0g of carbon, 53.3g of oxygen, and 6.7g of hydrogen. If we could count or otherwise determine the number of atoms ofeach element represented by these masses, we would have the molecular formula. However, this would not only be extremelydifficult to do but also unnecessary.

From our determination of atomic masses, we can note that 1 atom of carbon has a mass which is 12.0 times themass of a hydrogen atom. Therefore, the mass of $N$ atoms of carbon is also 12.0 times the mass of $N$ atoms of hydrogen atoms, no matter what $N$ is. If we consider this carefully, we discover that 12.0g of carbon contains exactly the same number of atoms as does 1.0g ofhydrogen. Similarly, we note that 1 atom of oxygen has a mass which is $\frac{16.0}{12.0}$ times the mass of a carbon atom. Therefore, the mass of $N$ atoms of oxygen is $\frac{16.0}{12.0}$ times the mass of $N$ atoms of carbon. Again, we can conclude that 16.0g of oxygen containsexactly the same number of atoms as 12.0g of carbon, which in turn is the same number of atoms as 1.0g of hydrogen. Without knowing (or necessarily even caring) what the number is, we can say that itis the same number for all three elements.

For convenience, then, we define the number of atoms in 12.0g of carbon to be 1 mole of atoms. Note that 1 mole is a specific number ofparticles, just like 1 dozen is a specific number, independent of what objects we are counting. The advantage to defining themole in this way is that it is easy to determine the number of moles of a substance we have, and knowing the number ofmoles is equivalent to counting the number of atoms (or molecules) in a sample. For example, 24.0g of carbon contains 2.0 moles ofatoms, 30.0g of carbon contains 2.5 moles of atoms, and in general, $x$ grams of carbon contains $\frac{x}{12.0}$ moles of atoms. Also, we recall that 16.0g of oxygen contains exactly as many atoms as does 12.0gof carbon, and therefore 16.0g of oxygen contains exactly 1.0 mole of oxygen atoms. Thus, 32.0g of oxygen contains 2.0 moles of oxygenatoms, 40.0g of oxygen contains 2.5 moles, and $x$ grams of oxygen contains $\frac{x}{16.0}$ moles of oxygen atoms. Even more generally, then, if we have $m$ grams of an element whose atomic mass is $M$ , the number of moles of atoms, $n$ , is

$n=\frac{m}{M}$

Now we can determine the relative numbers of atoms of carbon, oxygen, and hydrogen in our unknown compoundabove. In a 100.0g sample, we have 40.0g of carbon, 53.3g of oxygen, and 6.7g of hydrogen. The number of moles of atoms in eachelement is thus

${n}_{C}=\frac{40.0g}{12.0\frac{g}{\mathrm{mol}}}=3.33\mathrm{moles}$
${n}_{O}=\frac{53.3g}{16.0\frac{g}{\mathrm{mol}}}=3.33\mathrm{moles}$
${n}_{H}=\frac{6.7g}{1.0\frac{g}{\mathrm{mol}}}=6.67\mathrm{moles}$

We note that the numbers of moles of atoms of the elements are in the simple ratio ${n}_{C}:{n}_{O}:{n}_{H}=1:1:2$ . Since the number of particles in 1 mole is the same for all elements, then itmust also be true that the number of atoms of the elements are in the simple ratio 1 : 1 : 2. Therefore, the molecular formula of thecompound must be $CO{H}_{2}$ .

Or is it? On further reflection, we must realize that the simple ratio 1 : 1 : 2 need not represent the exactnumbers of atoms of each type in a molecule of the compound, since it is indeed only a ratio. Thus the molecular formula could just aseasily be ${C}_{2}{O}_{2}{H}_{4}$ or ${C}_{3}{O}_{3}{H}_{6}$ . Since the formula $CO{H}_{2}$ is based on empirical mass ratio data, we refer to this as the empirical formula of the compound. To determine the molecular formula , we need to determine the relative mass of a molecule of the compound,i.e. the molecular mass. One way to do so is based on the Law of Combining Volumes,Avogadro's Hypothesis, and the Ideal Gas Law . To illustrate, however, if we were to find that the relative mass of one molecule of the compound is60.0, we could conclude that the molecular formula is ${C}_{2}{O}_{2}{H}_{4}$ .

## Review and discussion questions

State the Law of Combining Volumes and provide an example of yourown construction which demonstrates this law.

Explain how the Law of Combining Volumes, combined with theAtomic-Molecular Theory, leads directly to Avogadro's Hypothesis that equal volumes of gas atequal temperatures and pressure contain equal numbers of particles.

Use Avogadro's Hypothesis to demonstrate that oxygen gasmolecules cannot be monatomic.

The density of water vapor at room temperature and atmospheric pressure is $0.737\frac{g}{L}$ . Compound A is 80.0% carbon by mass, and 20.0% hydrogen. Compound B is 83.3%carbon by mass and 16.7% hydrogen. The density of gaseous Compound A is $1.227\frac{g}{L}$ , and the density of Compound B is $2.948\frac{g}{L}$ . Show how these data can be used to determine the molar masses of Compounds A and B, assuming that water hasmolecular mass 18.

From the results above , determine the mass of carbon in a molecule of Compound A and in a molecule ofCompound B. Explain how these results indicate that a carbon atom has atomic mass 12.

Explain the utility of calculating the number of moles in a sample of asubstance.

Explain how we can conclude that 28g of nitrogen gas ( ${N}_{2}$ ) contains exactly as many molecules as 32g of oxygen gas ( ${O}_{2}$ ), even though we cannot possibly count this number.

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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Sherica
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Uday
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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Cied
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Porter
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Porter
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Cesar
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Stotaw
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Azam
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Damian
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Azam
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Prasenjit
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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