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# Elemental Analysis

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Have you ever played a prank where you put salt in a sugar container? Some poor unsuspecting soul will put two teaspoons of salt in their coffee and be immediately disgusted. It is common in chemistry, and life in general, for compounds to look similar. Sometimes cues like smell will clue you into the difference (alcohol smells very different from water), but that is only sometimes the case.

So how can we tell things apart if they aren't labeled? Chemists utilize a technique called elemental analysis to figure out the composition of compounds. Using this technique, it would be easy to distinguish between sugar and salt and avoid another ruined coffee! In this article, we will take a look at elemental analysis and all its uses within chemistry.

• First, we discuss the different applications of elemental analysis.
• Next, we cover its different methods.
• Then, we walk through a general procedure.
• Lastly, we learn about the formula used in elemental analysis and walk through an example.

## Applications of elemental analysis

Elemental Analysis is a process where a sample is analyzed for its elemental and sometimes isotopic composition.

Isotopes are different forms of the same element with the name number of protons, but a different number of neutrons

For isotopic composition, we would be determining the amount of different isotopes. For example, seeing the amount of carbon-12 versus carbon-14.

The first thing to think about regarding elemental analysis is understanding why you would use this technique. One important use would be for tissue and or blood samples. If you need to know what elements a tissue/blood sample was exposed to, then you would use elemental analysis to figure it out. In the discipline of Toxicology, elemental analysis is an often-used technique in analyzing substances.

Elemental analysis has wide applications in many scientific industries, such as pharmaceutical industrial work, research, geology, materials science, environmental science, and many more. If bulk analysis needs to be done, such as in the pharmaceutical industry, then the most common method would be X-ray fluorescence. Elemental analysis is a crucial analytical technique in many research fields and is used throughout the world as a premier way to analyze the elemental composition of substances.

## Elemental Analysis Methods

Now that we know why someone would use elemental analysis, let's dive into the how.

There are four standard methods of conducting elemental analysis using instruments:

1. X-ray fluorescence.
2. Atomic absorption.
3. Inductively coupled plasma - mass spectrometry.
4. inductively coupled plasma - optical emission spectroscopy.

Each of these analytical techniques has its own benefits, and all four are commonly used in labs throughout the world.

Mass spectrometry is the most common method to conduct elemental analysis (we will also be walking through an example later). Since it's the most common, we are going to dive a little deeper and walk through the steps:

1. A sample is ionized - usually an electron is lost to form a cation (positively charged ion).
2. The ions are sorted and separated based on their mass and charge.
3. The separate ions are measured, and the data is displayed on a chart.
• The data will have mass/charge (m/z) on the x-axis and relative abundance on the y-axis.

Below is an image of a mass spectrometer:

Fig. 1 - A mass spectrometer.

By performing mass spectrometry, results for the percent composition of each element are elucidated (which we will discuss later), which are used to determine the molecular chemical formula.

The molecular chemical formula gives us the exact ratio of elements. This formula is different from the empirical chemical formula, which tells us the smallest ratio of elements within a molecule. For example, the empirical formula would give us NO2, while the molecular formula is actually N2O4.

## Elemental Analysis Procedure

Now that we are familiar with the different methods, let's walk through how to use them. When conducting elemental analysis, there are two types of analysis: a qualitative and quantitative analysis.

Qualitative analysis focuses on determining what elements are in a sample, while quantitative analysis measures the amount of each element in the sample.

The general procedure for qualitative analysis is as follows:

1. Choose a method to use for the sample.
2. Using the chosen method, you obtain the identity of the elements.

The quantitative procedure is the same, except it measures the percent composition.

Percent composition is the total percentage of an element within a molecule.

So our steps would be:
1. Choose a method to use for the sample
2. Using the chosen method, obtain the percent composition
3. Using the percent composition, calculate the empirical or molecular formula (whichever is relevant for the method chosen)

## Elemental Analysis formula

For whichever method we use, it will spit out a chart showing the percent composition.

Using atomic absorption, the following percent composition data were obtained: 40.9 % Carbon (C), 4.57 % Hydrogen (H), and 54.5 % Oxygen (O). What is the empirical formula of the sample?

Our first step is to convert from a percentage to grams. We can assume that the total gram amount is 100 g. Technically, we can choose any amount since the ratio will be the same, but this method is simpler. Doing this conversion, we get:

$$100\,g*40.9\%=40.9\,g\,C$$

$$100\,g*4.57\%=4.57\,g\,H$$

$$100\,g*54.5\%=54.5\,g\,O$$

Next, we need to convert to moles. This is because the empirical formula (and all chemical formulas) are in units of moles. To do this, you divide the number of moles by each element's atomic mass, which is the amount of grams of the element per mole. It is also the weight listed under the element on the periodic table.

$$\frac{40.9\,g}{\frac{12.01\,g}{mol}}=3.41\,mol\,C$$

$$\frac{4.57\,g}{\frac{1.01\,g}{mol}}=4.52\,mol\,H$$

$$\frac{54.5\,g}{\frac{16.00\,g}{mol}}=3.41\,mol\,O$$

Our last step is to divide each molar amount by the smallest variable. Doing so will give us the ratio between elements.

$$\frac{3.41\,mol\,C}{3.41}=1\,C$$

$$\frac{4.52\,mol\,H}{3.41}=1.33\,H$$

$$\frac{3.41\,mol\,O}{3.41}=1\,O$$

We have a ratio of 1:1.33:1; however, chemical formulas are in whole numbers. So what do we do? Well, we multiply our ratio by the smallest number possible to get all whole numbers. In this case, that number is 3.

$$1\,C*3=3\,C$$

$$1.33\,H*3=4\,H$$

$$1\,O*3=3\,O$$

So our final formula is:

$$C_3H_4O_3$$

A key thing to note is that calculating the molecular formula is the same steps, except we would be using exact gram amounts and not estimated ones.

Let's walk through an example:

Using atomic absorption, the following percent composition data were obtained: 85.6 % Carbon (C) and 14.4 % Hydrogen (H). What is the molecular formula of the sample if the total mass is 84.16 g?

Since we know the actual mass, we will multiply the percentages by that mass:

$$84.16\,g*85.6\%=72.1\,g\,C$$

$$84.16\,g*14.4\%=12.1\,g\,H$$

Now we convert from grams to moles:

$$\frac{72.1\,g}{\frac{12.01\,g}{mol}}=6\,mol\,C$$

$$\frac{12.1\,g}{\frac{1.01\,g}{mol}}=12\,mol\,H$$

So the molecular formula is:

$$C_{6}H_{12}$$

## Elemental analysis example

Now that we've covered all of our basics, it's time for an example!

An unknown gaseous sample was analyzed using a mass spectrometer, giving the data below. What is the molecular formula of the sample?

Fig. 2 - Mass spectrometry data for an unknown sample.

Our first step is to understand what this data is telling us. Mass spectrometry breaks things into fragments, not necessarily elements, so each peak is a fragment or an element. The peak that has 100% intensity is the peak for the molecule itself, so we know that the entire molecule has a mass of 44 g.

This means that the other peaks are either elements or fragments (i.e., smaller molecules). Let's label each peak, going from left to right.

The first peak is 12 g, which means it is carbon, which has an atomic mass of 12 g/mol. The second peak is 16 g, so it is oxygen (oxygen has an atomic mass of 16 g/mol).

Now for the 28 g peak. It could be either silicon or carbon monoxide (CO), which both have masses of 28 g/mol. We know that this sample is a gas, so it is more likely that this is CO and not silicon (which is usually a solid). This also makes sense since the total mass of the sample is 44 g, and the mass of C+O+Si is 56 g.

Now that we know our peaks, we can determine the identity of the sample. So we know that the molecule contains two elements: C and O, so we can subtract those masses from the total mass.

$$44-(12.00+16.00)=16.00$$

Since there are 16 g "left over", that means there is another oxygen atom present, so the complete formula is $$CO_2$$

## Elemental Analysis - Key takeaways

• Elemental Analysis is a process where a sample is analyzed for its elemental and sometimes isotopic composition.
• The molecular chemical formula gives us the exact ratio of elements. This is different from the empirical chemical, which tells us the smallest ratio of elements within a molecule.
• Qualitative analysis focuses on determining what elements are in a sample, while quantitative analysis measures the amount of each element in the sample.
• Percent composition is the total percentage of an element within a molecule.
• The basic steps of elemental analysis are:
1. Choose a method to use for the sample.
2. Using the chosen method, obtain the percent composition.
3. Using the percent composition, calculate the empirical or molecular formula (whichever is relevant for the method chosen).

## References

1. Fig. 1 - A mass spectrometer (https://upload.wikimedia.org/wikipedia/commons/thumb/5/50/Thermo_-_Finnigan_LCQ_Mass_Spectrometer_%2815797493459%29.jpg/640px-Thermo_-_Finnigan_LCQ_Mass_Spectrometer_%2815797493459%29.jpg) by Kitmondo LAB (https://www.flickr.com/people/129143611@N03) licensed by CC BY 2.0 (https://creativecommons.org/licenses/by/2.0/)

## Frequently Asked Questions about Elemental Analysis

Elemental Analysis is a process where a sample is analyzed for its elemental and sometimes isotopic composition.

Elemental analysis is used to identify the elemental or isotopic composition of a sample. It is used in the fields of toxicology, research geology, environmental science, and many others.

From elemental analysis, we get the percent composition. We multiply each element's percent composition by the total gram amount to get the number of grams per element. Then, you divide each gram amount by the molar mass to get the amount of moles. You then divide each molar amount by the smallest molar amount to get the ratio between elements (the formula).

Organic compounds like blood and tissue are commonly analyzed using elemental analysis. The process is the same for inorganic and organic compounds.

There are 4 main methods of elemental analysis: X-ray fluorescence, atomic absorption, mass spectrometry, and optical emission spectroscopy.

## Elemental Analysis Quiz - Teste dein Wissen

Question

What is elemental analysis?

Elemental Analysis is a process where a sample is analyzed for its elemental and sometimes isotopic composition.

Show question

Question

Which of the following are methods of elemental analysis? (Select all that apply)

Atomic absorption

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Question

If Si2O4 is the molecular formula. which of the following is the empirical formula?

SiO2

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Question

What is percent composition?

Percent composition is the total percentage of an element within a molecule

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Question

True or false: Mass spectrometry separates a sample into its elements

False

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Question

Which of the following is a use for elemental analysis?

Determining what a tissue sample has been exposed to

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Question

Using atomic absorption on an unknown sample, the following data was obtained: 45.9% carbon (C), 3.21% hydrogen, and 50.89% bromine (Br), what is the empirical formula of the sample?

C6H5Br

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Question

Using elemental analysis, the following data was obtained: 3.25% hydrogen (H), 19.37% carbon (C), and 77.39% oxygen (O). What is the empirical formula of the sample?

H2CO3

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Question

Using elemental analysis, the following data was obtained: 90.51% carbon (C) and 9.49% hydrogen (H). If the total mass of the sample is 106.16 g, what is the molecular formula of the sample?

C8H10

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Question

Using elemental analysis, the following data was obtained: 46.29% carbon (C), 9.65% hydrogen (H), 19.16% nitrogen (N), and 21.89% oxygen (O). What is the molecular formula if the total mass of the sample is 146.19 g?

C6H14N2O2

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Question

An unknown gas sample was ran through a mass spectrometer. The following data was obtained: A peak at 14 m/z (5 rel. abundance), a peak at 15 m/z (10 rel. abundance), a peak at 16 m/z, (80 rel. abundance), and a peak at 17 m/z (100 rel. abundance). What is the formula of the unknown sample?

NH3

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Question

What is quantitative analysis?

Quantitative analysis measures the amount of each element in the sample.

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Question

What is qualitative analysis?

Qualitative analysis focuses on determining what element are in a sample

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Question

Fragments are separated by what in a mass spectrometer?

Mass and charge

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Question

If H2O is the empirical formula, which of the following are possible molecular formulas?

H6O3

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