Tag Archives: measurement

Mass Percent Example Problem

Ruby Gemstone

A ruby is a gemstone comprised mostly of aluminum oxide. Its color comes from the addition of chromium atoms to the crystal lattice. Creative Commons License

Mass percent or mass percent composition is a measurement of concentration. It is a measure of the ratio of the mass of one part of a molecule to the mass of the total molecule and expressed as a percentage.

This example problem shows how to determine the mass percent composition of each element of a molecule and determine which element makes up most of the molecule by mass.

Problem

Rubies and sapphires are gemstones where their crystal structure is mostly comprised of aluminum oxide, Al2O3. Find the mass % of aluminum and oxygen. Which element makes up most of the mass of the molecule?

Solution

The formula to calculate mass % is

where massA is the mass of the part you are interested in and massTOTAL is the total mass of the molecule.

Using a Periodic Table, we find

mass Al = 26.98 g/mol
mass O = 16.00 g/mol

There are two atoms of aluminum in a Al2O3 molecule, so

massAl = 2⋅26.98 g/mol = 53.96 g/mol

There are three atoms of oxygen:

massO = 3⋅16.00 g/mol = 48.00 g/mol

Add these together to get the total mass of Al2O3

massTOTAL = massAl + massO
massTOTAL = 53.96 g/mol + 48.00 g/mol
massTOTAL = 101.96 g/mol

Now we have all the information we need to find the mass % of each element. Let’s start with aluminum.

mass percent of aluminum

mass % Al = 0.53 ⋅ 100
mass % Al = 53%

Find mass % of oxygen.

mass percent of oxygen

mass % O = 0.47 ⋅ 100
mass % O = 47%

Answer

In aluminum oxide, Al2O3, aluminum accounts for 53% of the total mass and oxygen is 47% of the total mass. Even though the two elements make up nearly equal parts of the molecule, there is more aluminum by mass in one molecule of aluminum oxide.

Remember to check your answers in mass percent by adding up the individual parts together to see if you have 100%. For our example, 53% + 47% = 100%. Our answer checks out.

 

Concentration Units For Solutions

Beaker and FlaskChemistry is a science which deals a lot with solutions and mixtures. Knowing just how much of one thing is mixed in with a solution is an important thing to know. Chemists measure this by determining the concentration of the solution or mixture.

There are three terms that need to be defined in concentration discussions: solute, solvent and solution.

Solute: The dissolved substance added to the solution.
Solvent: The liquid that dissolves the solute.
Solution: The combination of solute and solvent.

The relationship between these three terms is expressed by many different concentration units. The unit you choose to use depends on how the solution is going to be used in your experiments. Common units include molarity, molality, and normality. Others are mass percent, mole fraction, formality and volume percent. Each unit is explained along with information about when to use them and the formulas needed to calculate the unit.

Molarity

Molarity is the most common concentration unit. It is a measure of the number of moles of solute in one liter of solution. Molarity measurements are denoted by the capital letter M with units of moles/Liter.

The formula for molarity (M) is

This shows the number of moles of solute dissolved in a liquid to make one liter of solution. Note the amount of solvent is unknown, just that you end up with a known volume of solution.

A 1 M solution will have one mole of solute per liter of solution.  100 mL will have 0.1 moles, 2L will have 2 moles, etc.

Molarity Example Problem

Molality

Molality is another commonly used concentration unit. Unlike molarity, molality is interested in the solvent used to make the solution.

Molality is a measure of the number of moles of solute dissolved per kilogram of solvent. This unit is denoted by the lower case letter m.

The formula for molality (m) is

molality formula

Molality is used when temperature is part of the reaction. The volume of a solution can change when temperature changes. These changes can be ignored if the concentration is based on mass of the solvent.

Molality Example Problem

Normality

Normality is a concentration unit seen more often in acid-base and electrochemistry solutions. It is denoted by the capital letter L with units of moles/L. Normality is more concerned with the chemically active part of the solution. For example, take two acid solutions, hydrochloric (HCl) acid and sulfuric (H2SO4) acid. A 1 M solution of HCl contains one mole of H+ ions and one mole of Cl ions where a 1 M solution of H2SO4 contains 2 moles of H+ ions and one mole of SO4 ions. The sulfuric acid produces twice the number of active H+ ions as the same concentration of HCl. Normality addresses this with the idea of chemical equivalent units. Equivalent units are the ratio of the number of moles of solute to the number of moles needed to produce 1 mole of the active ion. In our example, this ratio is 1:1 for HCl, both H+ and Cl ions so the equivalent unit for both ions is 1. For H2SO4, the ratio is 1:12 for H+ and 1:1 for SO4. The equivalent unit for H+ is 2 and 1 for SO4.

This number is used to calculate the normality of a solution using the formula

Note it is essentially the same as the molarity equation with the addition of equivalent units.
For our example, the 1 M solution of HCl would have a normality of 1 N for both H+ and Cl and the 1 M H2SO4 would have a normality of 2 N for H+ and 1 N for SO4.

Mass Percent, Parts per Million and Parts per Billion

Mass percent or mass percent composition is a measurement to show the percentage composition by mass of one part of a solution or mixture. It is most often represented by a % symbol.

The formula for mass percent is

where A is the part needed and the total is the total mass of the solution or mixture. If all the mass percent parts are added together, you should get 100%.

Mass Percent Example

If you think of mass percent as parts per hundred, you can make the leap to the units of parts per million (ppm) and parts per billion (ppb). These two units are used when the solute’s concentration is very small compared to the volume measured.

The formula for parts per million is

and parts per billion

Note the similarity between mass% and these two equations.

Volume Percent

Volume percent is a concentration unit used when mixing two liquids. When pouring two different liquids together, the new combined volume may not be equal to the sum of their initial volumes. Volume percent is used to show the ratio of the solute liquid to the total volume.

The formula is very similar to the mass percent, but uses volume in place of the mass. VolumeA is the volume of the solute liquid and the volumeTOTAL is the total volume of the mixture.

On a side note, v/v % measurements of alcohol and water are labelled commercially with the unit known as Proof. Proof is twice the v/v % measurement of ethanol in the beverage.

Mole Fraction

Mole fraction is the ratio of the number of moles of a single component of a solution to the total number of moles present in the solution.

Mole fractions are often used when discussing mixtures of gases or solids, but could be used in liquids. Mole fraction is denoted by the Greek letter chi, χ. The formula to calculate mole fraction is

Formality

Formality is a less common concentration unit. It appears to have the same definition as molarity with the formula:

Notice how the only difference between formality and molarity is the letters F and M. The difference is formality disregards what happens to the solute after it is added to the solution. For example, if you take 1 mole of NaCl and add it to 1 liter of water, most people would say you have a 1 M solution of NaCl. What you actually have is a 1 M solution of Na+ and Cl ions. Formality is used when it matters what happens to the solute in the solution. The above solution is a 1 F solution of NaCl.

In solutions where the solute does not dissociate, such as sugar in water, the molarity and formality are the same.

Today In Science History – October 4 – Mole

Chemistry Mole

This chemistry mole would be better suited as a mascot for Mole Day than the actual mole unit.

On October 4, 1971 the General Conference on Weights and Measures (Conférence Générale des Poids et Mesures, CGPM) officially added the mole as a base quantity of the SI units of measure.

According to Resolution 3 of the 14th meeting of the CGPM, a mole is the amount of a substance of a system which contains as many elementary entities as there are atoms in 0.012 kilograms of carbon 12.

The number of elementary particles in one mole is equal to Avogadro’s number. The current best value for that number is 6.02214129(27)×1023.

The CGPM also decided when a mole should be used. The elementary entities described in the definition must be specified and may be atoms, molecules, ions, electrons, other particles or specified groups of particles. The mole is supposed to be applied to these types of particles, but can easily applied to other objects. Some with humorous effect such as 6.022×1023 avocados = 1 guacamole.

The mole is also the only SI unit to have its own holiday. October 23 (10-23) is known as Mole Day to chemists. The more precise celebrants observe the holiday from 6:02 am to 6:02 pm.

Mole In Chemistry

Chemistry Mole

This chemistry mole would be better suited as a mascot for Mole Day than the actual mole unit.

Moles in chemistry have nothing to do with the small brown/gray yard pests that dig under your flowers and lawn or a tasty sauce on a Mexican dish. A mole in chemistry is a SI unit of measurement. The mole unit describes the amount or number of things. The standard mole is defined as the number of atoms in 12 grams of the carbon isotope, carbon-12.

1 mole = 6.02214129(27) x 1023.

More commonly, 1 mole (abbreviated mol) is equal to 6.022 x 1023. This number is known as Avogadro’s number. The mole has no units. It just describes the number of something.

1 mole of atoms = 6.022 x 1023 atoms
1 mole of water = 6.022 x 1023 water molecules
1 mole of moles = 6.022 x 1023 moles.

It is much easier to write 1 mole than 6.022 x 1023. The mole unit is a convenient means to convert between atoms and molecules and mass of those atoms or molecules. Moles are important enough to chemists that they celebrate Mole Day on October 23 (10-23).

Mole Example Problem #1: Find the mass of a single atom.

Question: What is the mass of single atom of nitrogen?

Solution:  The atomic mass of an element is the the mass in grams of one mole of that element. When we look at the periodic table, we see the atomic mass of nitrogen is 14.001 grams/mole.

This means 1 mole of nitrogen atoms has a mass of 14.001 grams.
1 mole of N atoms = 6.022 x 1023 N atoms = 14.001 grams.

Divide both sides of the equation by 6.022 x 1023 N atoms to get the mass of one nitrogen atom.

Mole Example Math Step 1

1 N atom = 2.325 x 10-23 g.

Answer: One nitrogen atom has a mass of 2.325 x 10-23 grams.

Mole Example Problem #2: Mass of a known number of molecules.

Question: What is the mass in grams of 5 billion water molecules?

Solution:
Step 1: Find the mass of one mole of water.
Water is H2O. Therefore, to find the mass of one mole of water, we need to know the mass of hydrogen and the mass of oxygen.
On our periodic table we see the mass of one mole of hydrogen is 1.001 g and the mass of one mole of oxygen is 16.00 g.
One mole of water has 2 moles of hydrogen and 1 mole of oxygen.
Mass of one mole of water = 2 (mass of hydrogen) + 1 mass of oxygen
Mass of one mole of water = 2(1.001 g) + 16.00 g = 18.002 g

Step 2: Find the mass of 5 billion (5 x 109) water molecules.
For this part, we use Avogadro’s number in a ratio.Mole Math 2Solve for x grams

mole math 3x grams = 1.49 x 10-13 grams

Answer: 5 billion water molecules has a mass of 1.49 x 10-13 grams.

Mole Example Problem #3: Finding the number of molecules in a given mass.

Question: How many water molecules in 15 grams of ice?

Solution:
Step 1: Find the mass of one mole of water.
We did this step in the second example. One mole of water is 18.002 grams.

Step 2: Use Avogadro’s number in a ratio.mole math 4Solve for x H2O moleculesmole math 5

x H2O molecules = 5.018 x 1023 H2O molecules

Answer: There are 5.018 x 1023 water molecules in 15 grams of ice.

Today In Science History – June 22

June 22 is an important day in the history of the standards of measurement used in science today. On June 22, 1799 the first official standard meter and kilogram were presented at the Archives de la République in Paris.

King Louis XVI charged leading scientists of the time to create a system of measurements that would lead to the creation of the metric system. The base measurements would all be based on reproducible physical standards. Other measurements would all be derived from the base units by powers of 10.

The meter was to be the standard of length. Two ideas were put forth for this standard. One was the length of a pendulum necessary to have a period of one half second. The other was one ten-millionth of the length of the Earth’s meridian along a quadrant, or one fourth the circumference of the Earth from North to South. Ultimately, the meridian length was chosen since the force of gravity varies from place to place which would alter the pendulum’s period. The official length of one meter would be 10-7 of the distance from equator to pole along the meridian passing through Paris.

The kilogram was originally supposed to be called a “grave” and set to be equal to the mass of one liter of water at the ice point (very close to today’s 1 kilogram), The French Revolution interrupted this standardization work and when the new government got involved, they preferred a standard unit called a “gramme” based on one cubic centimeter of water at the ice point. The gramme standard had one problem. It was too small amount of mass for trade. Typically, goods do not change hands on the gram scale of measurements. A larger unit of mass was needed. A compromise was reached where the gramme would still be used, but the official physical standard would be the size of the original “grave”. Since the gramme was 1/1000th the size of the gramme, it took 1,000 grammes to equal one of the new standard kilogramme. This is why the kilogram the only Metric standard to contain a prefix in its name.

The standards for meter and kilometer have changed over the years. A meter is now the distance light travels in a vacuum in 1/299792458 of a second. The kilogram is represented by a mass of platinum-iridium alloy kept at the Bureau International des Poids et Mesures (BIPM) in Paris.

Calculate Percent Error

Percent error is the percent difference between a measured and expected value. (Sherman Geronimo-Tan)

Percent error is the percent difference between a measured and expected value. (image: Sherman Geronimo-Tan)

Percent Error Definition

Percent error, sometimes referred to as percentage error, is an expression of the difference between a measured value and the known or accepted value. It is often used in science to report the difference between experimental values and expected values.

The formula for calculating percent error is:Percent Error Formula

Note: occasionally, it is useful to know if the error is positive or negative. If you need to know the positive or negative error, this is done by dropping the absolute value brackets in the formula. In most cases, absolute error is fine. For example, in experiments involving yields in chemical reactions, it is unlikely you will obtain more product than theoretically possible.

Steps to Calculate the Percent Error

  1. Subtract the accepted value from the experimental value.
  2. Take the absolute value of step 1
  3. Divide that answer by the accepted value.
  4. Multiply that answer by 100 and add the % symbol to express the answer as a percentage.

Now let’s try an example problem.

You are given a cube of pure copper. You measure the sides of the cube to find the volume and weigh it to find its mass. When you calculate the density using your measurements, you get 8.78 grams/cm3. Copper’s accepted density is 8.96 g/cm3. What is your percent error?

Solution:
experimental value = 8.78 g/cm3
accepted value = 8.96 g/cm3

Step 1: Subtract the accepted value from the experimental value.

8.96 g/cm3 – 8.78 g/cm3 = -0.18 g/cm3

Step 2: Take the absolute value of step 1

|-0.18 g/cm3| = 0.18 g/cm3

Step 3: Divide that answer by the accepted value.Percent Error Math 3

Step 4: Multiply that answer by 100 and add the % symbol to express the answer as a percentage.

0.02 x 100 = 2
2%

The percent error of your density calculation was 2%.

Significant Figures and Uncertainty

Ruler

All measurements have a degree of uncertainty. This ruler has several different levels of precision. Accuracy and precision depend on both the measuring tool and the person doing the measuring. Credit: Public Domain/Gowolves09

All measurements have some degree of uncertainty in their value. This can be caused by the skill of the measurement taker or the tool used.

For example, if you are in a chemistry lab and you need to add 8 mL of liquid to a beaker. You could just pour water straight into the beaker and quit when you think you hit 8 mL. The error associated with this measurement would mostly be associated to your skill. You could use a beaker with markings every 5 mL and get pretty close, give or take a couple mL. You could use a graduated cylinder with markings every tenth of a mL and reliably get measurements between 7.9 and 8.1 mL. Here we see how the uncertainty can be affected by the measuring tool.

Uncertainty is expressed in significant figures. The more significant figures in a measurement, the more precise the measurement. There are five basic rules dealing with significant figures.

  1. Non-zero digits are always significant.
  2. All zeros between other significant digits are significant.
  3. The most significant figure, also called most significant digit, is the leftmost non-zero digit. For example: in the number 0.00321, the most significant figure is the 3.
  4. The least significant figure, or least significant digit is the rightmost digit. In the number 54.321, the least significant figure is 1. Keep in mind, zero can be the least significant digit. For example, the zero in 4.320 is the least significant figure.
  5. Any zero digit to the right of the decimal point is significant.
    For example 2 has one significant digit, but 2.0 has two significant figures.
  6. If no decimal point is present, the rightmost non-zero digit is the least significant figure.

Quick Tip To Calculate Significant Figures.
Write the number in scientific notation. The numbers ahead of the multiplier are all significant.

Example: How many significant figures are in the following numbers?
a) 23,000
b) 0.000504
c) 240.05
d) 4.000

Write each number in scientific notation.
a) 2.3 x 103
b)5.04 x 10-4
c) 2.4005 x102
d) 4.000 x 101
Now we can easily count the digits ahead of the multiplier to get the number of significant figures.
a) 2 significant figures
b) 3 significant figures
c) 5 significant figures
d) 4 significant figures

Significant Figures and Uncertainty in Calculations

Once you have your measurement, you are most likely going to use it in a calculation of some sort. In a calculation, the uncertainty of the result is determined by the uncertainty of the measurements.

  • Addition and Subtraction

When measurements are used in addition and subtraction, the uncertainty is determined by the uncertainty of the least precise measurement, not by the number of significant figures.
Example: Add the following three measurements: 24.21 cm, 5.005 cm and 22 cm.
If you add them up, you get 51.215 m. The least precise measurement is the 22 cm measurement, so the answer should have the same precision.
The value of the calculation would be reported as 51 m.

  • Multiplication and Division

When measurements are used in multiplication and division, the number of significant figures in the result with be the same as the number with the smallest number of significant figures.
Example: Divide 35.105 grams by 35 mL.
If you just divide the two numbers, you get 1.003 g/mL. The value you would report depends on the measurement with the least significant figures. The first measurement has 5 significant and the second has only 2 significant figures.
The reported value would then be 1.0 g/mL

  • Losing Significant Figures

Significant figures can be ‘lost’ in a calculation. For example, if you have a beaker that weighs 75.206 grams and you add water until the weight is 75.844. The water would weigh the difference between these two values.
75.844 g – 75.206 g = 0.638 g
The final result only has 3 significant figures when both measurements had 5 significant figures.

  • Exact Numbers

Occasionally, a calculation involves a number with an exact value rather than an approximation. This occurs in calculations using conversion factors, pure numbers or physical constants. The significant figures of these numbers do not affect the end result. For example, if you were to find the average of 10.3 cm, 12.7 cm and 14.5 cm, you would add the three numbers together to get 37.5 cm. You would then divide this by 3 to get the average or 12.5 cm. Even though 3 only has one significant figure, your answer is still 12.5 cm.

The use and rules of significant figures in science and engineering is standard in any field. Measuring is a basic skill in science and everyone needs to work under the same rules. It is best to learn them early and keep them in mind in all your work.

More about Measurements:
What is the Difference Between Accuracy and Precision?