what is experimental value in chemistry

#"Error" = "|experimental value - accepted value|"#

The difference is usually expressed as percent error .

#"% error" = "|experimental value - accepted value|"/"experimental value" × 100 %#

For example, suppose that you did an experiment to determine the boiling point of water and got a value of 99.3 °C.

Your experimental value is 99.3 °C.

The theoretical value is 100.0 °C.

The experimental error is #"|99.3 °C - 100.0 °C| = 0.7 °C"#

The percent error is #"|99.3 °C - 100.0 °C|"/"100.0 °C" = "0.7 °C"/"100.0 °C" × 100% = 0.7 %#

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Percent difference: What to place on the denominator?

I'm genuinely confused on what to put on the denominator for the formula $$\%~\mathrm{difference} = \mathrm{\left|theoretical~-~experimental\right|\over theoretical}$$ or some other similar formula with different variable names that mean the same thing.

Specifically, I have a reading on the iron content per pill written on the bottle, and I have an amount that I calculated from doing some math with the experimental data. The problem is, I know that the calculated value is experimental, but the iron content per pill isn't really theoretical because it's known. Anyway, which should I put in the denominator ?

• experimental-chemistry

• $\begingroup$ Also, welcome to Chem.SE! :-) $\endgroup$ –  hBy2Py Commented Jun 16, 2016 at 0:58
• $\begingroup$ If it's %difference, shouldn't you multiply the RHS of your equation by 100? $\endgroup$ –  Carlos Gouveia Commented Jun 16, 2016 at 1:54

2 Answers 2

The experimental value is the one you measured/calculated yourself, from your own experiments. The theoretical value is the one that you obtained from some third party, to which you are comparing your experimental value.

So, the number on the bottle is the theoretical , and the number from your data and calculations is the experimental .

thus in your example x could be theoretical and z experimental values thus you have y=(x-z)/x = 1-z/x . The sigma's are the respective measured standard deviations. Once you have the total error you can decide how to write %difference +- error. (The absolute value does not matter here as the formula squares values)

• $\begingroup$ While this type of uncertainty propagation calculation is important to know how to carry out, I don't think it answers the OP's question. $\endgroup$ –  hBy2Py Commented Jun 17, 2016 at 23:24
• $\begingroup$ I added a comment that should answer this at the end of the first paragraph. If the OP is doing the calculation and is worried about precision then its not much more effort to calculate the errors properly. $\endgroup$ –  porphyrin Commented Jun 20, 2016 at 6:50

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Introduction to Experimental Chemistry

Welcome to the world of chemistry!

In the lecture courses [1] , we explain the conceptual framework by which chemists currently describe and explain matter and its interactions. However, ultimately, we are trying to explain what is going on in nature.

In doing so, a quote from Stephen Hawkings’ A Brief History of Time  comes to mind.

No matter how many times the results of experiments agree with some theory, you can never be sure that the next time the result will not contradict the theory. On the other hand, you can disprove a theory by finding even a single observation that disagrees with the predictions of the theory.

Like all other natural sciences, chemistry is fundamentally an experimental subject. While theorists and computational chemists help shape the field, whatever theories and models are developed must be able to explain experimental results. The interplay of theory (used to explain experimental results) and experiment (used to characterize matter and test theories) is critical in the pursuit of science.

Beyond the cutting edge, we rely on experimental results on a day-by-day basis. For example:

• If you look at a packet of food, you will find nutritional facts such as the amount of energy needed. These are typically determined using calorimetry .
• In biochemistry, we often need to determine the concentration of a protein to ensure that our experiment is working. This is typically done with absorption spectroscopy .

You will learn about both of these techniques in this course sequence.

The goals of this course are to:

• Laboratory safety.
• Keeping and using a laboratory notebook.  Proper documentation is critical in all laboratory settings, as well as in healthcare (charting).
• Mastery of techniques used to prepare and conduct experiments, as well as to use chemical instrumentation.
• Design appropriate procedures to perform experiments.
• Illustrate and further explore concepts that are presented in the general chemistry lecture sequence. [2] . We will reference Tro: Chemistry – Structures and Properties , 2nd Ed in our course this year as this is the textbook we will use in the lecture course in 2021-2022; however, any general chemistry textbook will suffice to accompany this course.

The most important part about the chemistry lab is to stay safe and learn about the chemistry. Remember, while you do the experiments, your instructor is the guide. Be sure to ask questions whenever necessary for safety or success! Good luck!

• CHEM-C 105/106, Principles of Chemistry I/II ↵
• CHEM-C 105/106, Principles of Chemistry I/II. ↵

IU East Experimental Chemistry Laboratory Manual Copyright © 2022 by Yu Kay Law is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License , except where otherwise noted.

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What is experimental value?

The experimental value is the value that you get in an experiment. The absolute value of the difference between the two values (the “error”) is your experimental error. The difference is usually expressed as percent error.

Table of Contents

How do you find the experimental value in chemistry?

For example, to calculate the experimental value for an experiment with results of 7.2, 7.2, 7.3, 7.5, 7.7, 7.8 and 7.9, add them all together first to arrive at a total value of 52.6 and then divide by the total number of trials – 7 in this case.

What is experimental value and theoretical value?

The experimental value is your calculated value, and the theoretical value is your known value. A percentage very close to zero means you are very close to your targeted value, which is good.

What is experimental value in percent error?

Percent error is the difference between a measured or experiment value and an accepted or known value, divided by the known value, multiplied by 100%. For many applications, percent error is always expressed as a positive value. The absolute value of the error is divided by an accepted value and given as a percent.

What is the difference between accepted value and experimental value?

The accepted value of a measurement is the true or correct value based on general agreement with a reliable reference. For aluminum, the accepted density is 2.70g/cm3. The experimental value of a measurement is the value that is measured during the experiment.

Why are theoretical and experimental values different?

This difference is due to three factors: the variation of the diffusion voltage, the nonzero electric field at the boundaries of the depletion region, and the contribution of electrons and holes. The exact values also disagree with the experimental results.

How do you find the experimental value of pi?

How do you compare theoretical and experimental results?

Experimental probability is based on the results of several trials or experiments. Theoretical probability is calculated by taking the number of favorable outcomes over the total number of outcomes. Experimental probability is calculated by taking the actual outcomes over the total number of trials.

How do you find the experimental value of Gravity?

What is experimental error?

Experimental error is the difference between a measured value and its true value. In other words, it is the inaccuracy or inaccuracies that stop us from seeing an absolutely correct measurement. Experimental error is very common and is to some degree inherent in every measurement.

What is the theoretical value?

Sometimes referred to as a fair or hypothetical value, a theoretical value is the estimated price of an option. Sometimes referred to as a fair or hypothetical value, a theoretical value is the estimated price of an option. The options pricing may have to do with buying, selling, or a combination of the two.

What is the meaning of accepted value?

In science, and most specifically chemistry, the accepted value denotes a value of a substance accepted by almost all scientists and the experimental value denotes the value of a substance’s properties found in a localized lab.

What is an acceptance value?

Accepted value is usually a number (or value) that is regarded as true by the general public, scientists, mathematicians, etc. It is often a term that is used in science, especially chemistry. It’s different from experimental value, which is the value yielded by a researcher or experimenter.

What is difference between accuracy and precision?

Accuracy is the degree of closeness to true value. Precision is the degree to which an instrument or process will repeat the same value. In other words, accuracy is the degree of veracity while precision is the degree of reproducibility.

Why are experimental values lower than theoretical chemistry?

The actual yield for an experimental reaction is not equal to the theoretical yield because of side reactions that occur as well as impurities in the container that might hinder the reaction.

What is the difference between experimental and theoretical probabilities?

Theoretical probability describes how likely an event is to occur. We know that a coin is equally likely to land heads or tails, so the theoretical probability of getting heads is 1/2. Experimental probability describes how frequently an event actually occurred in an experiment.

What is the difference between experimental and theoretical research?

A theory is usually expected to explain existing experimental results and to predict new results, while an experiment is usually expected to check the validity of existing theories and to gather data for modifying them.

How is pi calculated manually?

Once you’ve got the circumference and diameter, plug them into the formula π=c/d, where “π” is pi, “c” is circumference, and “d” is diameter. Just divide the circumference by the diameter to calculate pi!

What is experimental result?

The outcome of an experiment or set of experiments, including observations and primary, processed, and analyzed data. ( NCI Thesaurus)

How do you find the experimental value of g given the slope?

The constant g can be calculated by dividing 39.5 (4p ) by the slope: g = 39.5/slope.

What is the experimental value of acceleration due to gravity?

The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s.

What is experimental g?

In a psychology experiment, the experimental group (or experimental condition) refers to the group of participants who are exposed to the independent variable. These participants receive or are exposed to the treatment variable.

What are the 3 types of experimental errors?

There are three types of errors: systematic, random, and human error.

What are the 3 types of errors?

• (1) Systematic errors. With this type of error, the measured value is biased due to a specific cause.
• (2) Random errors. This type of error is caused by random circumstances during the measurement process.
• (3) Negligent errors.

What are experimental errors in chemistry?

Experimental error DOES refer to the uncertainty about the accuracy of the results of an experiment. There are two types of experimental errors in chemistry: (a) random errors (or indeterminate errors) (b) systematic errors (or determinate errors, or inherent errors)

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3.9: Determining a Chemical Formula from Experimental Data

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In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

Derivation of Molecular Formulas

Recall that empirical formulas are symbols representing the relative numbers of a compound’s elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.

Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass . As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as n :

\[\mathrm{\dfrac{molecular\: or\: molar\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}{empirical\: formula\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}= \mathit n\: formula\: units/molecule}\]

The molecular formula is then obtained by multiplying each subscript in the empirical formula by n , as shown by the generic empirical formula A x B y :

\[\mathrm{(A_xB_y)_n=A_{nx}B_{nx}}\]

For example, consider a covalent compound whose empirical formula is determined to be CH 2 O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:

\[\mathrm{\dfrac{180\:amu/molecule}{30\:\dfrac{amu}{formula\: unit}}=6\:formula\: units/molecule}\]

Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:

\[\ce{(CH2O)6}=\ce{C6H12O6}\]

Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.

Determination of the Molecular Formula for Nicotine

Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

Determining the molecular formula from the provided data will require comparison of the compound’s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound’s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:

\[\begin{alignat}{2} &\mathrm{(74.02\:g\: C)\left(\dfrac{1\:mol\: C}{12.01\:g\: C}\right)}&&= \:\mathrm{6.163\:mol\: C}\\ &\mathrm{(8.710\:g\: H)\left(\dfrac{1\:mol\: H}{1.01\:g\: H}\right)}&&= \:\mathrm{8.624\:mol\: H}\\ &\mathrm{(17.27\:g\: N)\left(\dfrac{1\:mol\: N}{14.01\:g\: N}\right)}&&= \:\mathrm{1.233\:mol\: N} \end{alignat}\]

Next, we calculate the molar ratios of these elements.

The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C 5 H 7 N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/mol formula unit.

We calculate the molar mass for nicotine from the given mass and molar amount of compound:

\[\mathrm{\dfrac{40.57\:g\: nicotine}{0.2500\:mol\: nicotine}=\dfrac{162.3\:g}{mol}} \nonumber \]

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

\[\mathrm{\dfrac{162.3\:g/mol}{81.13\:\dfrac{g}{formula\: unit}}=2\:formula\: units/molecule} \nonumber \]

Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

\[\ce{(C5H7N)6}=\ce{C10H14N2} \nonumber \]

Exercise \(\PageIndex{5}\)

What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?

C 8 H 10 N 4 O 2

The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound’s percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound’s empirical formula. The empirical formula mass of a covalent compound may be compared to the compound’s molecular or molar mass to derive a molecular formula.

Combustion Analysis

When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO 2 and the hydrogen to H 2 O (Figure \(\PageIndex{2}\)). The amount of carbon produced can be determined by measuring the amount of CO 2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO 2 produced by determining the increase in mass of the CO 2 trap. Likewise, we can determine the amount of H produced by the amount of H 2 O trapped by the magnesium perchlorate.

Figure \(\PageIndex{2}\): Combustion analysis apparatus

One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO 2 , H 2 O, N 2 , and SO 2 , respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in Figure \(\PageIndex{3}\) and a typical combustion analysis is illustrated in Examples \(\PageIndex{3}\) and \(\PageIndex{4}\).

Example \(\PageIndex{3}\): Combustion of Isopropyl Alcohol

What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO 2 and 0.306 grams of H 2 O?

From this information quantitate the amount of C and H in the sample.

\[ (0.561\; \cancel{g\; CO_2}) \left( \dfrac{1 \;mol\; CO_2}{44.0\; \cancel{g\;CO_2}}\right)=0.0128\; mol \; CO_2 \]

Since one mole of CO 2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO 2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this?

\[ (0.0128 \; \cancel{mol\; C}) \left( \dfrac{12.011\; g \; C}{1\; \cancel{mol\;C}}\right)=0.154\; g \; C \]

How about the hydrogen?

\[ (0.306 \; \cancel{g\; H_2O}) \left( \dfrac{1\; mol \; H_2O}{18.0\; \cancel{g \;H_2O}}\right)=0.017\; mol \; H_2O \]

Since one mole of H 2 O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H 2 O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample.

When we add our carbon and hydrogen together we get:

0.154 grams (C) + 0.034 grams (H) = 0.188 grams

But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol:

0.255 grams - 0.188 grams = 0.067 grams oxygen

This much oxygen is how many moles?

\[ (0.067 \; \cancel{g\; O}) \left( \dfrac{1\; mol \; O}{15.994\; \cancel{g \;O}}\right)=0.0042\; mol \; O \]

Overall therefore, we have:

• 0.0128 moles Carbon
• 0.0340 moles Hydrogen
• 0.0042 moles Oxygen

Divide by the smallest molar amount to normalize:

• C = 3.05 atoms
• H = 8.1 atoms

Within experimental error, the most likely empirical formula for propanol would be \(C_3H_8O\)

Example \(\PageIndex{4}\): Combustion of Naphalene

Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO 2 and 11.30 mg of H 2 O. Determine the empirical formula of naphthalene.

Given : mass of sample and mass of combustion products

Asked for : empirical formula

• Use the masses and molar masses of the combustion products, CO 2 and H 2 O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene.
• Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene.

A Upon combustion, 1 mol of CO 2 is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H 2 O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO 2 and H 2 O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:

\[ mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C} \]

\[ = 1.883 \times 10^{-2} \, g \, C \]

\[ mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H} \]

\[ = 1.264 \times 10^{-3} \, g \, H \]

B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:

\[ moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C \]

\[ moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H \]

Dividing each number by the number of moles of the element present in the smaller amount gives

\[H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250\]

Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C 1.25 H 1.0 . Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C 5 H 4 as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C 10 H 8 , which is consistent with our results.

Exercise \(\PageIndex{4}\)

• Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO 2 and 14.53 mg of H 2 O. Determine the empirical formula of xylene.
• The empirical formula of benzene is CH (its molecular formula is C 6 H 6 ). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO 2 and H 2 O will be produced?

The empirical formula is C 4 H 5 . (The molecular formula of xylene is actually C 8 H 10 .)

33.81 mg of CO 2 ; 6.92 mg of H 2 O ​​​​​​

Contributors and Attributions

Mike Blaber ( Florida State University )

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors.  Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).