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# Introduction to gravimetric analysis: Volatilization gravimetry

Introduction to volatilization gravimetry and precipitation gravimetry. An example using volatilization gravimetry to determine the purity of a metal hydrate mixture.

## What is gravimetric analysis?

Gravimetric analysis is a class of lab techniques used to determine the mass or concentration of a substance by measuring a change in mass. The chemical we are trying to quantify is sometimes called the analyte. We might use gravimetric analysis to answer questions such as:
• What is the concentration of the analyte in a solution?
• How pure is our sample? The sample here could be a solid or in solution.
There are 2 common types of gravimetric analysis. Both involve changing the phase of the analyte to separate it from the rest of a mixture, resulting in a change in mass. You might hear either or both of these methods being called gravimetric analysis as well as the more descriptive names below.
Drawing of Alice from Lewis Carroll's "Alice in Wonderland" holding a brown bottle labeled, "Drink Me."
It is generally not recommended to drink a mystery liquid! Maybe Alice could have used gravimetric analysis to figure out what is in the bottle. How might she check for the presence of soluble silver salts? Image of Alice from Wikimedia Commons, public domain
Volatilization gravimetry involves separating components of our mixture by heating or chemically decomposing the sample. The heating or chemical decomposition separates out any volatile compounds, which results in a change in mass that we can measure. We will go through a detailed example of volatilization gravimetry in the next section of this article!
Precipitation gravimetry uses a precipitation reaction to separate one or more parts of a solution by incorporating it into a solid. The phase change occurs since the analyte starts in the solution phase and then reacts to form a solid precipitate. The solid can be separated from the liquid components by filtration. The mass of the solid can be used to calculate the amount or concentration of the ionic compounds in solution.
In this article, we will go through an example of using volatilization gravimetry in a chemistry lab setting. We will also discuss some of the things that might go wrong during a gravimetric analysis experiment and how that might affect our results.

## Example: Determining the purity of a metal hydrate mixture using volatilization gravimetry

Bad news! We have just been informed by our inept lab assistant, Igor, that he may have accidentally contaminated a bottle of the metal hydrate start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text with an unknown amount of start text, K, C, l, end text. In order to find the purity of our start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, we heat 9, point, 51, start text, g, end text of the metal hydrate mixture to remove water from the sample. After heating, the sample has a reduced mass of 9, point, 14, start text, g, end text.
What is the mass percent of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text in the original mixture?
Gravimetric analysis problems are simply stoichiometry problems with a few extra steps. Hopefully you will remember that in order to do any stoichiometric calculations, we need the coefficients from the balanced chemical equation.
First let’s analyze what is happening when we heat the sample. We are removing water from the start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text to form anhydrous start text, B, a, C, l, end text, start subscript, 2, end subscript, left parenthesis, s, right parenthesis and water vapor, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, g, right parenthesis. By the end of our heating process we should be left with a mixture of anhydrous start text, B, a, C, l, end text, start subscript, 2, end subscript, left parenthesis, s, right parenthesis and start text, K, C, l, end text, left parenthesis, s, right parenthesis. In the following calculations, we are going to make the following assumptions:
• All mass lost from the sample is from evaporated start text, H, end text, start subscript, 2, end subscript, start text, O, end text, as opposed to some other decomposition process.
• All of the water is coming from the dehydration of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text.
Note: We don’t know anything about how much of the contaminant, start text, K, C, l, end text, is in the mixture. It could be anywhere from 0, minus, 100, percent, start text, K, C, l, end text by mass! It probably isn’t 100, percent, start text, K, C, l, end text though, since we did lose water after heating.
We can express the dehydration reaction in terms of a balanced chemical equation:
start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, s, right parenthesis, right arrow, start text, B, a, C, l, end text, start subscript, 2, end subscript, left parenthesis, s, right parenthesis, plus, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, g, right parenthesis
Based on the above balanced equation, we expect to make 2, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, g, right parenthesis for every 1, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text. We will use this stoichiometric relationship in our calculations to convert the moles of water lost to the moles of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text in the original sample.
Let's go through the calculation step-by-step.

### Step $1$1: Calculate change in sample mass

We can find the amount of water lost during the heating process by calculating the change in mass for our sample.
\begin{aligned}\text{Mass of } \text H_2 \text O &= \text{Initial sample mass} - \text{Final sample mass} \\ &= 9.51\,\text{g}-9.14\,\text{g} \\ &=0.37\,\text{g H}_2 \text O \end{aligned}

### Step $2$2. Convert mass of evaporated water to moles

In order to convert the amount of water lost to the amount of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text using the mole ratio, we will need to convert the mass of evaporated water to moles. We can do this conversion using the molecular weight of water, 18, point, 02, start text, g, slash, m, o, l, end text.
start text, M, a, s, s, space, o, f, space, w, a, t, e, r, end text, equals, 0, point, 37, start cancel, start text, g, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, times, start fraction, 1, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, divided by, 18, point, 02, start cancel, start text, g, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, end fraction, equals, 2, point, 05, times, 10, start superscript, minus, 2, end superscript, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text

### Step $3$3. Convert moles of water to moles of $\text{BaCl}_2 \cdot 2 \text H_2 \text O$start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text

We can convert the moles of water to moles of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text using the mole ratio from the balanced reaction.
start text, m, o, l, space, o, f, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, equals, 2, point, 05, times, 10, start superscript, minus, 2, end superscript, start cancel, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, times, start fraction, 1, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, divided by, 2, start cancel, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, end fraction, equals, 1, point, 03, times, 10, start superscript, minus, 2, end superscript, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text

### Step $4$4. Convert moles of $\text{BaCl}_2 \cdot 2 \text H_2 \text O$start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text to mass in grams

Since we want to find the mass percent of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, we will need to know the mass of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text in the original sample. We can convert moles of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text to mass in grams using the molecular weight of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text.
start text, M, a, s, s, space, o, f, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, equals, 1, point, 03, times, 10, start superscript, minus, 2, end superscript, start cancel, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, times, start fraction, 244, point, 47, start text, g, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, divided by, 1, start cancel, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, end fraction, equals, 2, point, 51, start text, g, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text

### Step $5$5. Calculate mass percent of $\text{BaCl}_2 \cdot 2 \text H_2 \text O$start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text in the original sample

The mass percent can be calculated using the ratio of the mass from Step 4 and the original sample mass.
$\text{Mass% BaCl}_2 \cdot 2 \text H_2 \text O=\dfrac{2.51\,\cancel{\text g} \text{BaCl}_2 \cdot 2 \text H_2 \text O}{9.51\,\cancel{\text g} \text{of mixture}}\times 100\%=26.4\% \,\text{BaCl}_2 \cdot 2 \text H_2 \text O~~~~~~~\text{(No thanks to Igor!)}$
Shortcut: We could also combine steps 2 to 4 into a single calculation (with the caveat that we should pay extra close attention to our units). In order to convert the mass of start text, H, end text, start subscript, 2, end subscript, start text, O, end text to mass of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text (which I will call "hydrate" in the calculation to save some space), we could solve the following expression:
start text, M, a, s, s, space, o, f, space, h, y, d, r, a, t, e, end text, space, equals, space, start underbrace, 0, point, 37, start cancel, start text, g, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, space, times, space, start fraction, 1, start cancel, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, divided by, 18, point, 02, start cancel, start text, g, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, end fraction, end underbrace, space, times, space, start underbrace, start fraction, 1, start cancel, start text, m, o, l, space, h, y, d, r, a, t, e, end text, end cancel, divided by, 2, start cancel, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, end fraction, end underbrace, space, times, space, start underbrace, start fraction, 244, point, 47, start text, g, space, h, y, d, r, a, t, e, end text, divided by, 1, start cancel, start text, m, o, l, space, h, y, d, r, a, t, e, end text, end cancel, end fraction, end underbrace, space, equals, space, 2, point, 51, start text, g, space, h, y, d, r, a, t, e, end text space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, S, t, e, p, space, 2, colon, end text, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, S, t, e, p, space, 3, colon, end text, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, S, t, e, p, space, 4, colon, end text
space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, f, i, n, d, space, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, u, s, e, space, m, o, l, e, space, r, a, t, i, o, end text, space, space, space, space, space, space, space, start text, f, i, n, d, space, g, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space

## Potential sources of error

We just successfully used gravimetric analysis to calculate the purity of a mixture, hooray! However, sometimes when you are in lab, things might not go quite so smoothly. Things that could go wrong include:
• Stoichiometry errors, such as not balancing the equation for the dehydration of  start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text
• Lab errors, such as not giving the water enough time to evaporate or forgetting to tare a piece of glassware
What would happen to our answer for the above situations?
Situation 1: We forgot to balance the equation
In this situation, we would end up using the wrong mole ratio in the calculation in Step 3. Instead of using the correct ratio of start fraction, 1, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, divided by, 2, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end fraction, we would use the ratio start fraction, 1, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, divided by, 1, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end fraction. That would double the moles of metal hydrate calculated in Step 3, which will also double our overall mass percent of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text. Ultimately, we would end up concluding that our sample has a much higher purity than it actually does!
Concept check: What mass of metal hydrate would we calculate in situation 1?
The moral of this story? Double check that all equations are properly balanced!
Situation 2: We ran out of time and not all the water evaporated
Hand holding watchglass with white anhydrous copper(II) sulfate, and hydrated copper(II) sulfate, which appears as a sky blue spot in the middle of the white powder after water was added.
In some cases, the color differs between the metal hydrate and the anhydrous compound. For example, anhydrous copper(II) sulfate is a white solid that turns bright sky blue when it is hydrated. In such cases, you could use the color change as well as the mass to monitor the dehydration process. Image by Benjah-bmm27 on Wikimedia Commons, Public domain
In the second situation, we did not fully dehydrate our sample. This could happen for a lot of reasons, unfortunately. For example, we could run out of time, the heat could be set too low, or maybe we just took the sample off the heat before it was done by mistake. How does that affect our calculations?
In this situation, the difference in mass we calculate in Step 1 will be lower than it should be, so we will have correspondingly fewer moles of water in Step 2. That will result in calculating a lower percent mass of start text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text compared to fully dehydrating the sample. In the end, we will end up underestimating the purity of the metal hydrate.
Chemists usually try to avoid situation 2 by drying to constant mass. That means monitoring the change in mass during the drying period until you no longer observe any further change in mass (which also depends on the accuracy of your lab balance). When you first start heating your sample, you might expect to see a significant mass decrease as water is lost. As you continue to heat the sample, the change in mass gets smaller since there is less water left in the sample to evaporate. At some point, there won’t be enough water left to make a significant change in mass, so the measured mass will stay approximately constant over multiple measurements. At that point, you can hopefully assume your sample is dry!
Lab tip: Surface area is always a factor when removing volatiles from a sample. Having a higher surface area will increase the rate of evaporation. You can increase the surface area of the sample by spreading your sample as thinly as possible on the heating surface or breaking up any larger chunks of solid, since moisture can get trapped inside the chunks.

## Summary

Gravimetric analysis is a class of lab techniques that uses changes in mass to calculate the amount or concentration of an analyte. One type of gravimetric analysis is called volatilization gravimetry, which measures the change in mass after removing volatile compounds. An example of volatilization gravimetry would be using the change in mass after heating to calculate the amount or purity of a metal hydrate. Some useful tips for gravimetric analysis experiments and calculations are:
• Double check stoichiometry and make sure equations are balanced.
• When removing volatiles from a sample, make sure to dry to constant mass.
• Always tare your glassware!
To read more about another common type of gravimetric analysis, see this article on precipitation gravimetry.

## Want to join the conversation?

• What's the dot in BaCl​2​​⋅2H​2​​O supposed to mean? I've never seen an equation like this before...
• The dot shows that a crystal of BaCl₂ contains 2 molecules of water for every formula unit of BaCl₂.
The BaCl₂·2H₂O is called a hydrate, and the H₂O is called the water of hydration.
• what does taring a piece of glassware mean? Is it related to balancing the weight of a container?
• Imagine trying to weigh some jellybeans in a glass jar. If you placed them on a scale you would get the weight of the Jellybeans + the weight of the Jar. If you wanted to just find out the weight of the Jellybeans, you would weigh the Jar when empty, then weigh the Jellybeans and the Jar together, calculate the difference between the two measurements and you'd get the weight of just the Jellybeans. e.g
weight of Jar without Jellybeans = 100g
weight of Jar + Jellybeans = 400g
weight of Jellybeans = 400g - 100g = 300g
Measuring the weight of the empty container (in this case the Jar) is what Taring is... Taring isn't limited to jellybeans in glass jars though, it's really common in lots of chemistry stuff...
• I'm having a really hard time learning from these write-ups. I can read the same paragraph over and over but it just doesn't make sense to me. Videos I can typically watch on 1.5X speed and fully grasp the concept. Is there anything I can do to improve my ability for learning from write-ups?
• I think that if you rush, it can easily get really confusing... I think you should try to take it slow, read more slowly and carefully. Comprehend the meaning of each sentence before moving on to the next. You know?
• In Step 4, molecular weight for BaCl2.2H2O is 244.47g, that is when, the H2O is multiplied by the two in front. But in Step 2, to calculate the mass of water, the molecular weight used was 18.02, but, in the products side of the equation, the coefficient 2 is there infront of H2O. So why do we have to take into account the 2 in BaCl2.2H2O but not the one in 2H2O?
• The equation for the reaction is
BaCl₂⋅2H2O → BaCl₂+2H₂O
It tells you that 1 mol of BaCl₂⋅2H₂O forms 2 mol of H₂O.
In step 4, you are finding the mass of 1 mol of BaCl₂⋅2H₂O.
1 mol of BaCl₂⋅2H₂O contains 1 formula unit of BaCl₂ and 2 mol of H₂O
∴ MM of BaCl₂⋅2H₂O = MM of BaCl₂ + 2 × MM of H₂O = 208.23 g + 2 × 18.016 g = 208.23 g + 36.03 g = 244.26 g

In step 3 and the step 5 shortcut, you are accounting for the 2 in front of H₂O.
It is in the mole ratio (1 mol hydrate/2 mol H₂O).
• You stated that KCl wasn't included in the equation because it is not going through physical or chemical reaction. How do we know that?
• It didn't said that KCl is not going through chemical reaction, instead we analyzed that KCl didn't reacted in the reaction and it was like spectator reactant. So rather that including this in our reaction and confusing ourselfs further we just cancelled it from both sides of the reaction since it didn't change at all ( didn't go through reaction).
• what is the purpose of gravimetric analysis
• A simple example would be to weigh a tree limb before and after baking it to determine the water content of it.
• What is the purpose behind finding the mass % of Bacl2 * 2H2O?

Aren't we curious how much KCL igor put in the solution? What does knowing that the mass % is 26.38 tell us?

Let's say the new solution with KCL now contains 100g of liquid.

Does that mean 26.38g of the solution is BaCl2 and water? and the other 73.6 g is KCL?
• In the beginning of the example problem, BaCl2 and 2H2O are separated by a dot rather than a plus sign. Why is that, and what is the different?

I'm also confused about how we got those stoichiometric coefficients. The explanations says we expect to make 2 moles of H2O (g) for every one mole of BaCl2 [dot] 2H2O. But shouldn't it be 2 moles of H2O (g) for everyone one mole of BaCl2 and 2 moles of H2O (g) for every 2 moles of H2O from the reactant? I'm not sure if that makes a difference, and I apologize if my question in unclear, but hopefully someone will understand what I'm asking and be able to provide me an explanation.