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empirical formula calculator
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From Wikipedia
Empirical formula
In chemistry, the empirical formula of a chemical compound is the simplest whole number ratio of atoms of each element present in a compound. An empirical formula makes no reference to isomerism, structure, or absolute number of atoms. The empirical formula is used as standard for most ionic compounds, such as CaCl2, and for macromolecules, such as SiO2. The term empirical refers to the process of elemental analysis, a technique of analytical chemistry used to determine the relative amounts of each element in a chemical compound.
In contrast, the molecular formula identifies the number of each type of atom in a molecule, and the structural formula also shows the structure of the molecule.
For example, the chemical compound n-hexane has the structural formula CH|3|CH|2|CH|2|CH|2|CH|2|CH|3, which shows that it has 6 carbon atoms arranged in a straight chain, and 14 hydrogen atoms. Hexane's molecular formula is C|6|H|14, and its empirical formula is C|3|H|7, showing a C:H ratio of 3:7. Different compounds can have the same empirical formula. For example, formaldehyde, acetic acid and glucose have the same empirical formula, CH|2|O. This is the actual chemical formula for formaldehyde, but acetic acid has double the number of atoms and glucose has six times the number of atoms.
Examples of common substances
Use in physics
In physics, an empirical formula is a mathematical equation that predicts observed results, but is derived from experiment or conjecture and not directly from first principles.
An example was the Rydberg formula to predict the wavelengths of hydrogen spectral lines. Proposed in 1888, it perfectly predicted the wavelengths of the Lyman series, but lacked a theoretical basis until Niels Bohr produced his Bohr model of the atom in 1913.
From Yahoo Answers
Question:Combustion analysis of fluorene, a polycyclic aromatic hydrocarbon used to make dyes, plastics, and pesticides, produced 11.44g CO2 and 1.80g H20
fluorene only has hydrogen and carbon in it.
calculate the empirical formula for fluorene.
will you include how you got this please?
Answers:H2O has a formula mass of 18.0153 and you generated 1.80g of it. 1.80g H2O / (18.0153 g H2O / mole H2O) = 0.0999 moles H2O Since 1 mole of H2O containd 2 moles of H, you had originally 0.1998 moles of H. CO2 has a formula mass of 43.9992 and you generated 11.44g of it. 11.44g CO2 / (43.9992 g CO2 / mole CO2) = 0.2600 moles of CO2 Since 1 mole of CO2 contains 1 mole of C, you had originally 0.2600 moles of C You must now find a whole number ratio from these. Change these to 19.98 for H and 26.00 for C. Now round 19.98 to 20.00 for H. Divide by 2 to get C13H10 If you google Fluorene, formula you will find its formula to be C13H10. Remember, with empirical formulas, it could have just as easily been a multiple of this (ie C26H20, or C39H30). C13H10
Answers:H2O has a formula mass of 18.0153 and you generated 1.80g of it. 1.80g H2O / (18.0153 g H2O / mole H2O) = 0.0999 moles H2O Since 1 mole of H2O containd 2 moles of H, you had originally 0.1998 moles of H. CO2 has a formula mass of 43.9992 and you generated 11.44g of it. 11.44g CO2 / (43.9992 g CO2 / mole CO2) = 0.2600 moles of CO2 Since 1 mole of CO2 contains 1 mole of C, you had originally 0.2600 moles of C You must now find a whole number ratio from these. Change these to 19.98 for H and 26.00 for C. Now round 19.98 to 20.00 for H. Divide by 2 to get C13H10 If you google Fluorene, formula you will find its formula to be C13H10. Remember, with empirical formulas, it could have just as easily been a multiple of this (ie C26H20, or C39H30). C13H10
Question:Epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress, contains 59.0 %C, 7.1%H, 26.2%O, and 7.7N% by mass; its MW is about 180 amu .
Answers:Assume you have 100 grams, so each percentage becomes grams. Divide each grams by the respective mol. mass of the atom. Divide the resulting moles by the smallest number. This should give you whole numbers for the EF Calculate the EF mol. mass. Divide 180 by this number to see how many units of the EF there are in the MF
Answers:Assume you have 100 grams, so each percentage becomes grams. Divide each grams by the respective mol. mass of the atom. Divide the resulting moles by the smallest number. This should give you whole numbers for the EF Calculate the EF mol. mass. Divide 180 by this number to see how many units of the EF there are in the MF
Question:Adipic acid is an organic compound composed of 49.31% C, 43.79% O, and the rest hydrogen. If the molar mass of adipic acid is 146.1 g/mol, what are the empirical and molecular formulas for adipic acid? (Type your answer using the the order C H O or formatted as CO2 for CO2.)
Answers:The trick to this question is knowing what to do about the percentages they give you. The most that we can have of the compound is 100.00%, so add the two percents given and you have the percent of H. Now, the trick. Let's assume that we have 100.00 g of the acid. This means that: There are 49.31 g of C, 43.79 g of O, and 6.90 g of H. Now divide each of these numbers by their respective molar masses (from the periodic table) to get them all on the same scale of moles. Yes, chemists are obsessed with moles!!! :) C: 49.31 g C x (1 mol C/12.01 g C) = 4.106 mol C O: 43.79 g O x (1 mol O/16.00 g O) = 2.737 mol O H: 6.90 g H x (1 mol H/1.008 g H) = 6.845 mol H Now, divide all of these moles by the smallest number obtained, in this case 2.737 mol. Thus, 4.106 mol C/ 2.737 mol = 1.500 C 2.737 mol O/ 2.737 mol = 1.000 O 6.845 mol H/ 2.737 mol = 2.501 H If we round these to 2 decimal places, we get the ratio of C:O:H to be 1.5:1.0:2.5 The empirical formula at this point looks like C1.5 O1.0 H2.5. We need whole numbers in there, so look at the formula. If we double everything, we will get nice round answers, so we get C3O2H5 This is the empirical formula (although the question wants it like C3H5O2). Now, how do we get the molecular formula? First, we calculate what the "empirical weight" is. We do that by using the molar masses from the periodic table and multiplying them by the ratios of the formula. C: 3 x 12.01 g/mol = 36.03 g/mol O: 2 x 16.00 g/mol = 32.00 g/mol H: 5 x 1.008 g/mol = 5.04 g/mol Summing these up, we get 73.08 g/mol. This is the "empirical weight." The molecular mass was given to us, 146.1 g/mol. If we divide this by the empirical mass, we get 146.1 g/mol / 73.08 g/mol = 1.999 or 2. This is the factor we need to multiply our empircal formula by to get the molecular formula. Thus, our molecular formula is: C6O4H10, or C6H10O4 for the format the question is asking you. Hope this helps.
Answers:The trick to this question is knowing what to do about the percentages they give you. The most that we can have of the compound is 100.00%, so add the two percents given and you have the percent of H. Now, the trick. Let's assume that we have 100.00 g of the acid. This means that: There are 49.31 g of C, 43.79 g of O, and 6.90 g of H. Now divide each of these numbers by their respective molar masses (from the periodic table) to get them all on the same scale of moles. Yes, chemists are obsessed with moles!!! :) C: 49.31 g C x (1 mol C/12.01 g C) = 4.106 mol C O: 43.79 g O x (1 mol O/16.00 g O) = 2.737 mol O H: 6.90 g H x (1 mol H/1.008 g H) = 6.845 mol H Now, divide all of these moles by the smallest number obtained, in this case 2.737 mol. Thus, 4.106 mol C/ 2.737 mol = 1.500 C 2.737 mol O/ 2.737 mol = 1.000 O 6.845 mol H/ 2.737 mol = 2.501 H If we round these to 2 decimal places, we get the ratio of C:O:H to be 1.5:1.0:2.5 The empirical formula at this point looks like C1.5 O1.0 H2.5. We need whole numbers in there, so look at the formula. If we double everything, we will get nice round answers, so we get C3O2H5 This is the empirical formula (although the question wants it like C3H5O2). Now, how do we get the molecular formula? First, we calculate what the "empirical weight" is. We do that by using the molar masses from the periodic table and multiplying them by the ratios of the formula. C: 3 x 12.01 g/mol = 36.03 g/mol O: 2 x 16.00 g/mol = 32.00 g/mol H: 5 x 1.008 g/mol = 5.04 g/mol Summing these up, we get 73.08 g/mol. This is the "empirical weight." The molecular mass was given to us, 146.1 g/mol. If we divide this by the empirical mass, we get 146.1 g/mol / 73.08 g/mol = 1.999 or 2. This is the factor we need to multiply our empircal formula by to get the molecular formula. Thus, our molecular formula is: C6O4H10, or C6H10O4 for the format the question is asking you. Hope this helps.
Question:Guys, please help me out. I've been racking my brain for about 20 minutes just for this single problem, and I have wayyyy too much stuff to do.
Terephthalic acid is composed of C, H, and O. When .6943g of tetephtalic acid is subjected to combustion analysis, it produced 1.471 g CO2 and .225 g H2O. What is it's empirical formula?
I know, combustion is
terephathalic acid + O2--> CO2 + H2O
but i don't understand how everything connects.
Explain, please?
Answers:Let us start from basic principles. You already know that if you combust terephthalic acid you produce CO2 and water. You have provided an equation to show this You also know that a certain mass 1.471g CO2 was produced. The C in the CO2 was originally the C in terephthalic acid. We know enough about CO2 to be able to calculate the exact mass of C in 1.471g CO2. This then is the mass of C in the original 0.6943g terephthalic acid. We can do the same with the H2O - we canm calculate the h in the H2O, which gives us the mass of H in the original sample of terephthalic acid. We know C, we know H , so by difference we can calculkate O in the original sample. Let us do this: Mass C in 1.471g CO2 Molar mass C = 12.011g/mol Molar mass CO2 = 44.009g/mol Mass C = 12.011/44.009*1.471 = 0.40146g C in sample Mass H in 0.225g H2O Molar mass H = 1.008, H2 = 2.016g/mol Molar mass H2O = 18.015g/mol Mass H = 2.016/18.015* 0.225 = 0.02518g Mass O = 0.6943 - ( 0.40146+0.02518) = 0.26766g Now divide each mass by respective atomic mass: C = 0.40146/12.011 = 0.033 H = 0.02518/1.008 = 0.025 O = 0.26766/15.999 = 0.01673 Divide through by smallest: C = 0.033/0.01673 = 1.97 = 2 H = 0.025/0.01673 = 1.5 O = 0.01673/0.01673 = 1.0 To remove fractions: multiply by 2: C = 4 H= 3 O=2 Empirical formula = C4H3O2
Answers:Let us start from basic principles. You already know that if you combust terephthalic acid you produce CO2 and water. You have provided an equation to show this You also know that a certain mass 1.471g CO2 was produced. The C in the CO2 was originally the C in terephthalic acid. We know enough about CO2 to be able to calculate the exact mass of C in 1.471g CO2. This then is the mass of C in the original 0.6943g terephthalic acid. We can do the same with the H2O - we canm calculate the h in the H2O, which gives us the mass of H in the original sample of terephthalic acid. We know C, we know H , so by difference we can calculkate O in the original sample. Let us do this: Mass C in 1.471g CO2 Molar mass C = 12.011g/mol Molar mass CO2 = 44.009g/mol Mass C = 12.011/44.009*1.471 = 0.40146g C in sample Mass H in 0.225g H2O Molar mass H = 1.008, H2 = 2.016g/mol Molar mass H2O = 18.015g/mol Mass H = 2.016/18.015* 0.225 = 0.02518g Mass O = 0.6943 - ( 0.40146+0.02518) = 0.26766g Now divide each mass by respective atomic mass: C = 0.40146/12.011 = 0.033 H = 0.02518/1.008 = 0.025 O = 0.26766/15.999 = 0.01673 Divide through by smallest: C = 0.033/0.01673 = 1.97 = 2 H = 0.025/0.01673 = 1.5 O = 0.01673/0.01673 = 1.0 To remove fractions: multiply by 2: C = 4 H= 3 O=2 Empirical formula = C4H3O2
From Youtube
Molecular and Empirical Formulas :Introduction to molecular and empirical formulas. Calculating molecular mass.
How to Calculate Empirical Formula from Mass Data | www.whitwellhigh.com :How to Calculate Empirical Formula from Mass Data Chemistry Whitwell High School UTC - University of Tennessee at Chattanooga www.whitwellhigh.com Instructor/Professor: Johnny Cantrell
