Empirical Formula Vitamin C Information

About Empirical Formula Vitamin C

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How do you calculate empirical/molecular formula of vitamin C (molar mass =176.12g/mol)? Vitamin C (molar mass = 176.12 g/mol) is a compound of C, H and O only. When a 2.000g sample was burned it produced 3.00 g of carbon dioxide and 0.82 g of water. Calculate the empirical formula and the molecular formula of vitamin C. I know how to calculate the number of moles of CO2 and H2O but then I don't know where to go from there.... can someone please help?!? THANK YOU SO MUCH!

Natalie Z replied: "It looks like you have Scerri for Chem 20A right? Anyways, me and my roommate figured it out last night. If you take a look at the example in the book (6th Edition) in Ch2 for the empirical formula in regards to combustion, they've basically showed you half the equation. If you take the 3g of CO2/44.01g/mol CO2, you get .0682 mol. If you take the .82g of H2O/18.02g/mol H2O, you get .0455 mol but you have to double that because it's 2 H molecules to one O molecule, so you get .0910 mol of H Now you just convert the mol to g. .0682 mol x 12.01 g C = .8187g C .0910 mol x 1.008 g H = .0918g H 2g Vitamin C - (.8187 g C + .0918g H) = X g of O X = 1.0895g O Convert O to mol (1.0895 g O/15.9994 g = .0681 mol O) If you look at the ratio between O and H, it's basically a 3:4 ratio. So the empirical formula becomes C3H4O3. To get the molecular formula: Molar mass of C3H4O3 = 3(12.01) + 4(1.008) + 3(16) = 88.067 Take the original molar mass of Vitamin C (176.12) and divide it by 88.062 and you get 2. Therefore, you double the empirical formula to get C6H8O6 as the molecular formula."

Vitamin C has an empirical formula of C3H4O3 and a molecular mass of 176 amu. What is its molecular formula? Can anyone please help, I have no idea how to do this so step by step would be nice.

nemeczek replied: "calculate MW of C3H4O3 divide 176 by MW (you should get 2) Multiply C3H4O3 by the number you got above"

Charles replied: "What you have to do is make the amu of each element, multiplied by its subscript, equal 176. So, to keep the ratio the same, you have to multiply every subscript by the same number. The equation is 176= x(36+4+48) 176= 88x 2=x Two is the magic number we're looking for... Multiply each element's subscript by 2, and the answer is C6H8O6"

The molecular formula for vitamin C is C6H8O6. What is the empirical formula? a.CHOc.C3H4O3 b.CH2Od.C2H4O2

ChemTeam replied: "c"

Stylish Nerd! replied: "option c is the right choice"

lenpol7 replied: "c !!!!"

Gagan replied: "C3H4O3.obtained by dividing by 2 to get simplest ratio."

Finding empirical formula with specific grams.? Vitamin C contains the elements of carbon, hydrogen and oxygen. A student analyzes a 5.00 g tablet of vitamin C and found it contains 40.9% carbon and 4.58% hydrogen by mass. What is the empirical formula of Vitamin C. I am not sure if i can just assume a 100 g sample like other empirical formula problems are solved. If you could explain that would really help! THANKS!

Andrew replied: "C3H4O3 percent of element = atomic mass x number of atoms / molar mass of compound x 100 carbon: 40.9 = 12 x #of atoms / molar mass of compound x 100 40.9 = 1200 x number of atoms / molar mass of cmpd. (174) 40.9 = 6.90 x number of atoms divide both by 6.9 5.9 = number of atoms so its really 6 so now we have C6H_O_ hydrogen: 4.58 = 1 x #of atoms / molar mass of compound x 100 4.58 = 100 x number of atoms / molar mass of cmpd. (174) 4.58 = .57 x number of atoms divide by .57 8 = the number of atoms now its C6H8O_ oxygen: 54.52 = 16 x #of atoms / molar mass of compound x 100 54.52 = 1600 x number of atoms / molar mass of cmpd. (174) 54.52 = 9.2 x number of atoms divide by 9.2 5.9 = number of atoms so 6 so the molecular formula for vitamin C is C6H8O6 which make the empirical C3H4O3"

ceddog86 replied: "As long as the mass is expressed in percent, you can assume a 100 g sample. If they gave you the mass of the whole product and the mass of the different elements in the sample, you would have to calculate the percentage first. In this question the mass of the original sample of vitamin C is extra information. From the way that you asked the question it sounds like you know how to solve these. Briefly, assume a 100 g sample of vitamin C. It contains 40.9 g carbon and 4.58 g hydrogen. Use the atomic mass to calculate the number of moles of each contained in the given number of grams. 40.9 g C * (1 mol / 12 g) = moles of carbon 4.58 g H * (1 mol / 1 g) = moles of carbon The rest is made up of oxygen. The percent mass oxygen is: 100 - (40.9 + 4.58) = 54.52% oxygen by mass. 54.52 g O * (1 mol / 16 g) = moles of oxygen."

ascorbic acid (vitamin acid) contains 40.92% C, 4.58% H, 54.50% O by mass. what is the empirical formula of? ascorbic acid?

Doc89891 replied: "Divide the percentages by atomic masses and form small whole numbers 40.92/12 = 3.41 4.58/2 = 2.29 54.5/16 = 3.406 now divide each answer by the smallest answer 3.41/2.29 = 1.489 2.29/2.29 = 1.0000 3.406/2.29 = 1.487 C and O come out at 1.5 compared to the 1 - so double everything to make C and O come out 3 2.98 2.00 2.97 so your empirical formula is C3H2O3"

empirical help!? Vitamin C has an Empirical formula of C3H4O3 and a molecular mass of 176 amu. What is its molecular formula?

Wander replied: "C3H4O3 = 176 g C3H4O3?"

cihanmccabe replied: "C3H4O3"

Joe replied: "Add the total formula mass of C3H4O3 and you get: 12*3 + 1*4 + 16*3 = 88 amu Divide 176 by 88 and you get 2, therefore you need to double the empirical formula, and get: C6H8O6"

chemistryyyyyy helppppp? 1) A hydrocarbon (an organic compound con- taining only C and H) is found to be 7.74% H. If the molecular weight is known to be be- tween 70 and 85 u, find the molecular formula. 2) A compound is analyzed and found to con- tain 36.70% potassium, 33.27% chlorine, and 30.03% oxygen. What is the empirical for- mula of the compound? 3) A pure sample contains only nitrogen and oxygen atoms. If the sample is 30.4% nitro- gen, by weight, what is the empirical formula of the molecule? 4) A sample of vitamin C was analyzed and found to contain 40.9% carbon, 4.58% hydro- gen, and 54.5% oxygen. What is the empirical formula for vitamin C? 5) A substance has a molar mass of 60 g/mol with 40.0 percent carbon, 6.7 percent hydro- gen, and 53.3 percent oxygen. What is its empirical formula? 6) A compound consists of 65.45% C, 5.492% H, and 29.06% O on a mass basis and has a molar mass of 110 g/mol. Determine the molecular formula of the compound. 7) Identify the empirical formula of C6H12O6

kitesurfah3792 replied: "Here's an example of how to do these from my old Chem 1 class notes: Ex. Calculate the empirical formula of the compound with the following % composition: 79.8% C 20.2 % H 1. erase % and replace with g. (because 1 mol = gam, gmm, gfm) 2. Find number of mol and divide the number of mol of each element by the lowest number. 3. The whole number = number of atoms of each element. 79.8 g C x 1 mol C = 6.65 mol C/ 6.65 = 1 12 g C 20.2 g H x 1 mol H = 20.2 mol H/6.65 = 3 1 g H Answer: CH3 Just follow those steps, and it'll work out. For the 1st one, to get %oxygen, do 100% - %Hydrogen"

chemistry problems? 1.what is the chemical test to differentiate saturated and unsaturated fats? 2.what is the different between the molecular formula & empirical formula of vitamin c? 3.what is the condition necessary for hydrogenation? 4. how is it possible for hydrogenation to be carried out?

Merlin's Feline replied: "One of the simplest tests used for unstauration in fatty acids is the iodine test wherein a solution of iodine is decolorized by reaction with unsaturated fatty acids...the more unsaturation the more the iodine solution is decolorized .It is the basis of the iodine number for unsat/sat 2. a molecular formula is the actual formula based on the true molecular weight of a compound .The empirical formula is merely the simplest atom to atom ratio comprising the percent composition. So for vitamin C the relationship between the atoms is C3 H4 O3 but the molecular weight is TWICW what the empirical formula reveals so the true molecular formula is C6 H8O The necessary condition for hydrogenation is that a compound exists that can add hydrogen under conditions that achieve or exceed the energy of activation for the process. Usually a double bond or groups of douyble bonds are most attractive for this purpose. ( for instance converting polyunstaurated oils to monounsaturated , more solid derivatives Usually for catalytic hydrogenation the compound inplced in a hydrogen atmosphere ( sometimes under pressure ) with a catalyst that brings the hydrogen and the compound's double bonds in appropriate association. The Hydrogen adda across the double bond to yield a saturated compound H2/Pd-C R-CH=CH- CH2 COOH ------ R-CH2CH2CH2-COOH"