In the ICSE Chemistry exam, a question on empirical formula had percentage compositions adding up to 99.9% but had no effect on the answer. I only wrote the question number. Will I get bonus marks for a wrong question?
HiIf u have attempted the question correctly and got the perfect emperical formula that is NH3 then u would score full in that question but if u have just wrote the question number and not solved further or solved it incorrectly , u will get no bonus marks
Chem question. empirical formula.?
Ibuprofen, the active ingredient in Advil, is made up of carbon, hydrogen, and oxygen atoms. When a sample of ibuprofen, weighing 5.00 g, burns in oxygen, 13.86 g of CO2 and 3.926 g of water are obtained. What is the simplest formula [empirical formula] of ibuprofen??? thanks ♥
Empirical formula question! Chemistry?
Divide each mass value by respective atomic mass Zn = 9.75/65 = 0.15 P = 3.1/31 = 0.1 Divide through by smaller value Zn = 0.15/0.1 = 1.5 P = 0.1/0.1 = 1 Bring to whole numbers by multiplying through by 2 Zn = 1.5*2 = 3 P = 1*2 = 2 Empirical formula = Zn3P2 % P = 3.1/(9.75+3.1) * 100 %P = 3.1/12.85*100 %P = 24.1% P
Kind of hard empirical formula questions..chemistry?
I will help you on Problem #1, and you should solve Problem#2 by yourself with the same principle I will show you. OK? You need to look for the molar mass of the following: CO2: 44.01 g/mol H2O: 18.015 g/mol C: 12.01 g/mol H: 1.008 g/mol O: 16.00 g/mol Therefore, you may understand that 44.01g CO2 contains 12.01g of C, and so on and so forth. Thus, 5.607 g of CO2 contains 5.607g*(12.01/44.01) of C, or 1.530 g of C. 1.148g of H2O contains 1.148g*(2.016/18.015) of H, or 0.128g of H. That is to say, the 2.338 g of this substance contains 1.530 g of C and 0.128g of H, and therefore must contain this much O: 2.338g - 1.530g - 0.128g = 0.680g (of O) 1.530 g of C = 0.127 mole of C 0.128g of H = 0.127 mole H 0.680g of O = 0.0425 mole O Notice that 0.127 is about 3 times of 0.0425. The empirical formula of the compound is C3H3O. Now, please follow my method to solve #2.
How to do empirical formulas in chemistry GCSE?
Hi,Empirical Formulae are very typical type of questions often asked in Exams. It is the most often asked questions in GCSE as well as A-levels.Empirical Formulae tell you the simple ratio of all the atoms present in the compound. Do not confuse this with Molecular Formulae. Molecular Formulae tell you the exact number of each atom present in the compound. For example, Glucose Molecule has the molecular formulae as C6H12O6 which means one molecule of glucose contains 6 atoms of carbon, 12 atoms of hydrogen and 6 atoms of oxygen. However, this is not the ratio of the atoms. The empirical formulae of this molecule will be CH2O.Now these are the steps that you should follow for empirical formulae:Make the following table and start filling the same.Now lets try to fill this with an example. If suppose we have a question:A sample of a compound is found to contain 63.5% silver, 8.2% nitrogen and 28.3% oxygen. What is the empirical formula for this compound?We have have three elements in the above question. Ag, N and O. Now lets fill the table with the data above.Hence the Empirical Fomulae for this compound will be:AgNO3I hope it answers your questions. If you need additional worksheets or help on this topic then you can visit www.expertguidance.co.uk
Why is the empirical formula used in chemistry?
In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.[1]A simple example of this concept is that the empirical formula of sulfur monoxide, or SO, would simply be SO, as is the empirical formula of disulfur dioxide, S2O2.This means that sulfur monoxide and disulfur dioxide, both compounds of sulfur and oxygen, will have the same empirical formula.However, their chemical formulas, which express the number of atoms in each molecule of a chemical compound, may not be the same.An empirical formula makes no mention of the arrangement or number of atoms. It is standard for many ionic compounds, like calcium chloride (CaCl2), and for macromolecules, such as silicon dioxide (SiO2).The molecular formula, on the other hand, shows the number of each type of atom in a molecule. The structural formula shows the arrangement of the molecule. It is also possible for different types of compounds to have equal empirical formulas.Samples are analyzed in specific elemental analysis tests to determine what percent of a particular element the sample is composed of.ref.-Wikipedia, the free encyclopedia
Chemistry- Empirical Formula Problem?
mass of Fe = 0.424 moles of Fe = 0.424 / 55.84 = 0.0076 moles mass of O = .606 - .424 g = 0.182 g and 0.182 / 16 moles = 0.0114 molar ratio of Fe : O = 0.0076: 0.0114 divide by the smallest number and we get 1: 1.5 x 2 we have 2:3 empirical formula is Fe2O3
A chemistry problem, empirical formula?
CeCl3 moles AgCl = 1.15g / 143.32 g/mol moles AgCl = 0.008024 moles gram Cl = 0.008024 mol * 35.453 g/mol gram Cl = 0.284 grams gram Ce = 0.660g - 0.284g gram Ce = 0.376 grams moles Ce = 0.376g / 140.12g/mol moles Ce = 0.002683 moles ratio of Ce and Cl, Ce = 0.002683 mol / 0.002683 mol = 1 Cl = 0.008024 mol / 0.002683 mol = 3 hence, empirical formula is CeCl3
Difficult Chemistry Question: Finding an Empirical Formula?
A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced 0.213 g CO2 and 0.0310 g H2O. In another experiment, it is found that 0.103 g of the compound produces 0.0230 g NH3. What is the empirical formula of the compound? Hint: Combustion involves reacting with excess O2. Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O. Also assume that all the nitrogen ends up in the NH3 in the second experiment. According to my periodic table, the molar masses of the elements involved are as follows: C: 12.011 g/mol H: 1.00794 g/mol O: 15.9994 g/mol N: 14.0067 g/mol My initial answer was C7H5N2O7, but this is wrong. There was another question just like this on Answers, but the only person who answered that one said it was C7H5O6N3, which is also wrong. I would greatly appreciate help from the first person to show me how to get to the correct empirical formula.
Can you explain empirical formula? (chemistry)?
You might get this easy thru examples.Look at this empirical formula: C2H3O All of the following then will have that empirical formula: C4H6O2 --subscript were each multiplied by 2 C6H9O3subscript were each multiplied by 3 C8H12O4 C10H15O5 It is silimlar to saying that the numbers 4,10,6,8,20,24 are multiples of 2. In this case, 2 is the "empirical formula" of these numbers