Percent by mass of carbon in a 50g sample of C2H2? (check work)?
If I want to determine the percent by mass of carbon in a 50 g sample of C2H2, how would I go about doing that? I know how to determine the amount of carbon, but I'm not sure where the 50g comes into play, if at all. Forgive me if this is a stupid question, but I haven't slept in awhile. %C = (2)(12) amu/26 amu * 100 = 92.3% C Thanks!
A 0.1375 g sample of solid magnesium is burned in a constant volume bomb calorimeter that has a heat capacity?
Total heat: (3024 J/C)*(1.126C) = 3.405 kJ For every gram of Mg: 3.405 kJ/0.1375g = 24.76 kJ/g Since the molar mass of Mg is: 24.305g/mol, the molar heat is: (24.76 kJ/g)*(24.305g/mol) = 601.9 kJ/mol. Please study hard -- you are asking too simple questions.
How many carbon atoms are present in 22g of CO2?
The process is simple for finding the number of carbon atoms in 22 grams of carbon dioxide. At first, we will find the number of moles present in given mass of carbon dioxide and then multiply it by Avogadro constant as well as with the number of moles of carbon present independent in one molecule of carbon dioxide.The process can be carried out as shown in the picture below.
A 57 g sample of iron reacts with 24 g of oxygen to form how many grams of iron oxide?
4 Fe + 3 O2 = 2 Fe2O3 moles Fe = 57 g/ 55.847 g/mol=1.02 moles O2 required = 1.02 x 3/4=0.765 actual moles O2 = 24 g/ 32 g/mol=0.75 => limiting reactant moles Fe2O3 = 0.75 x 2/3 =0.50 mass Fe2O3 = 0.50 mol x 159.694 g/mol=80 g
What is the percent yield for this reaction?
CaCO3(s) ⟶ CaO(s)+CO2(s) moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 from bal. rxn. 1 mole of CO2 forms per mole CaCO3, therefore 0.0131 moles CO2 should form. 0.0131 moles CO2 x 44 g/mole CO2=0.576 g CO2 % Yield: 0.53/.576 x100= 92 percent yield
0.290 g of an organic compound containing C,H and O gave on combution 0.270 g of water and 0.66 g of CO2. What is empirical formula of the compound?
First calculate %of C, H AND O by using following formula% C = [12/44][mass of carbon dioxide/mass of oragic compound] * 100= 62.07%% of H =[2/18][mass of water/mass of organic compound]*100 = 10.34% of O = 100 - (10.34 +62.07) = 27.59Next you need to follow the following stepsCalculate moles of each element as uou know about mass from the above mass percent valuesNo. of moles ofC =5.1725. H = 10.34. O = 1.72Divide with number of moles by least number value i.e., 1.72 you get simple whole number ratioC = 3. H = 6. O = 1Empirical formula is C3H6O
How many carbon atoms in a 3-carat pure diamond, which has a mass of 0.60 grams?
According to Wikipedia one Mol of diamond weighs 12.01 grams. We know that one Mol of any substance contains 6.02 x 10^23 molecules. That means that 3 carat (0.6 gram) diamond contains (6.02 x 10^23 x 0.6) / 12.01 molecules. That amounts to about 3 x 10^22 molecules. In case of diamond, the number of atoms is the same as the number of molecules.So it's a 3 with 22 zeroes, 30 sextillion or in other words it's thirty thousand billion billion atoms - quite a big number. It's about 30000 times more than the number of stars in the observable universe.
What is the molecular formula of a gaseous hydrocarbon which contains 82.76% of carbon and the vapour density is 29?
The molecular formula of a compound represents the actual number of atoms of various elements present in a molecule of a compound.Whereas, Empirical formula represents the simplest whole number ratio in which the atoms are present in a molecule of that compound.In order to derive molecular formula of any unknown compound, we need to know the following relations:Molecular formula = n * Empirical formula, where n = molecular formula mass/empirical formula massMolecular mass = 2 * Vapor densityNow we know the percentage of carbon in the hydrocarbon which is 82.76% so we can assume the rest of the element is hydrogen only and then its percentage will be 17.24%.Considering 100 g of the samplemass of carbon = 82.76 gmass of hydrogen = 17.24 gWe can convert these masses in terms of moles by dividing with respective molar massesNumber of moles of carbon = 82.76/12 = 6.8 molesNumber of moles of hydrogen = 17.24/1 = 17.24 molesWe need to convert these number of moles in terms of simple whole number ratio, we can divide both by smaller one in order to find it. Hence,6.8/6.8 = 117.24/6.8 = 2.5We multiply both by 2 to convert them in simple whole numbersCarbon=2 and hydrogen = 5Thus Empirical formula of the compound is given by, C2H5Empirical formula mass of C2H5 = 2*12 + 5*1 = 29 gNow we can determine molecular formula mass from vapor density = 2 * 28 = 58 gTherefore n = 58/29 = 2Thus Molecular formula of the compound = 2 * C2H5 = C4H10
On complete combustion, 0.246 g of an organic compound gave 0.198 g 0f CO2 and 0.1014 g of H2O. What is the ratio of carbon and hydrogen atoms in the compound?
Given:mass of organic compound --- 0.246 gmass of CO2 formed ----- 0.198 gpercentage of Carbon = 12 x mass of CO2 formed x 100 44 mass of organic compoundpercentage of carbon = 12 /44 x 0.198/0.246 x 100percentage of carbon = 2.376/10.824 x 100% carbon = 0.2195 x 100%carbon = 21.95----------------------------------------------------------21.95% of carbon is present in the compoundPercentage of hydrogen =2/18× mass of H2O/mass of compound×100 = 2×0.1014×100/18×0.246 = 4.58Hence,the composition of compounds areCARBON : 21.95%HYDROGEN : 4.58%.All the best if u guys get more doubts ask me i will help u out of it.Your friend,GRENINJA.