A sample of urea contains 1.233 g N, 0.177 g H, 0.528 g C, and 0.704 g O. What is the empirical formula of urea?

Q: A 125.6 mg sample of an organic compound containing only carbon, hydrogen, and nitrogen is combusted…

A: Answer The empirical formula is C3H8N The molecular formula is C6H16N2

Q: An Organic compound contains 59.00% by mass C, 7.15% by mass H, 7.65% by mass N, and 26.20% by mass…

A: Given data is as follows: The elements are given in percent which is out of 100. So, the mass of C =…

Q: Malonic acid is used in the manufacture of barbiturates (sleeping pills). The composition of the…

A: To calculate the mole of each elements present in Malonic acid: Given: 34.56 % of C, 3.9 % of H, and…

Q: The compound chloral hydrate, known in detective stories as knockout drops, is composed of 14.52% C,…

A: The empirical formula of a compound gives the proportions of the elements present in the compound.…

Q: Combustion analysis of a hydrocarbon produced 33.01 g CO2 and 4.83 g H2O. What is the empirical…

A: Given : Mass of CO2 produced = 33.01 g Mass of H2O produced = 4.83 g And the compound is…

Q: A compound is composed of carbon, hydrogen, and oxygen. When 10.0 g of an unknown compound was…

A: First Calculate mole of carbon ,hydrogen then Calculate mass of oxygen and mole of oxygen. Then…

Q: A compound was found to contain 115.49g of Ca, 92.16 g of O, and 5.82 g of H. What is the total mass…

A: mass of Ca = 115.49g mass of O = 92.16 g mass of H =  5.82 g  Molar mass of Ca = 40.078 g/molMolar…

Q: An organic compound has an empirical formula CH2 and a molar mass of 112.21 g/mol. How many H atoms…

A: an empirical formula CH2  empirical mass = (12 x 1 ) + (1 x 2)= 14  n = molar mass / empirical mass…

Q: A compound contains only C, H, and N. Combustion of 35.0 mg of the compound produces 54.9 mg CO2 and…

A:

Q: How would you find the yu of water (in mJ/m²) if you have a n-Heptane hydrocarbon with a y of…

A: A cubic yard is 3′ x 3′ x 3′. 3 x 3 x 3 = cubic feet for every cubic yard = CFPCY New water…

Q: Combustion analysis of 2.400 g of an unknown compound containing carbon, hydrogen, and oxygen…

A: Solution                       We know that Number of moles = (Given Mass) / (Molecular Mass) Mass…

Q: Vitamin C is essential for the prevention of scurvy (and large doses may be effective in preventing…

A:

Q: Opal is a hydrated form of silica. If a laboratory analysis of a sample of opal reveals it to…

A: Hydrated form of silica means some of water molecules are present in form of H2O in silica compound.

Q: Zinc phosphate is used as a dental cement. A 50.00-mg sample is broken down into its constituent…

A:

Q: A 1.000 g sample of an unknown compound containing only carbon, hydrogen, and oxygen was analyzed…

A: Given: Mass of sample = 1.000 g Mass of CO2 formed = 2.579 g Mass of H₂O formed = 0.6536 g Molar…

Q: A 1.000 g sample of an unknown compound containing only carbon, hydrogen, and oxygen was analyzed…

A:

Q: A compound contains 78.14% B and 21.86% H by mass and has a molar mass of 27.67 g/mol. What is the…

A: mass % of B = 78.14 %mass % of H = 21.86 % mass of hydrogen in 1 mole (27.67 g) of the compound = ?

Q: A 5.17 g sample of a compound with the formula CxHyOz is combusted and yields 13.0 g CO2 and 2.29 g…

A:

Q: A 125.6 mg sample of an organic compound containing only carbon, hydrogen, and nitrogen is combusted…

A:

Q: Vitamin C is essential for the prevention of scurvy (and large doses may be effective in preventing…

A:

Q: One medication used to treat diabetic kidney disease, high blood pressure, and heart failure is made…

A:

Q: A 4.73 mg sample of a compound containing carbon, hydrogen, and oxygen underwent combustion…

A:

Q: Polyethylene is a polymer consisting of only carbon and hydrogen. If 2.300 g of the polymer is…

A:

Q: 3) The combustion of 4.000 g of a compound that contains only C, H, N, and Br yields 3.826 g of CO2…

A:

Q: A given compound is analyzed and found to contain only C, H and O. When a 35.79 g sample of this…

A: Given,

Q: The composition of a compound was found to be 66.7% C, 25.9% N and 7.41% H by mass? What is the…

A: Empirical formula represnts number of atoms of various elements present in a molecule in simple…

Q: A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields…

A:

Q: Complete combustion of a 0.500 g sample of a pure hydrocarbon yielded 0.973 g CO 2 and 0.319 g H 2O.…

A:

Q: A molecule has an empirical formula of CH 2O and a molar mass of 179.9 g/mol. What is the molecular…

A: The empirical formula describes the smallest whole number ratio of atoms present in a compound.…

Q: experiment, 3.15 g of this compound  was combusted to produce 5.69 g of CO2 , 2.72 g H2O and 0.734 g…

A: Moles of C = 5.69/44 = 0.129 mol Mass of C = 0.129×12 = 1.55 g Moles of H = 2.72×2/18 = 0.302 mol…

Q: Combustion analysis of 2.796 g of an unknown compound containing carbon, hydrogen, and oxygen…

A:

Q: The empirical formula of styrene is CH; the molar mass of styrene is 104.14 g/mol. What number of H…

A:

Q: Complete combustion of 3.703.70 g of a hydrocarbon produced 12.012.0 g of CO2 and 3.703.70 g of H2O.…

A:

Q: A 4.78 g sample of a compound with the formula CyH,Oz is combusted and yields 11.9 g CO2 and 5.86 g…

A: Empirical formula of a compound is the formula which is represented by the ratio of minimum number…

Q: Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen, and oxygen produced…

A: Empirical formula is the simplest whole number ratio of the compound. Mass of sample = 1.031g Mass…

Q: What is the empirical formula for a compound that is 41.8% N, 24.7% C, 31.4% O, and 2.1% H by mass?…

A:

Q: Complete combustion of 8.582 g of a particular hydrocarbon yielded 27.73 g of CO2 and 9.080 g of…

A: Given,mass of the hydrocarbon = 8.582 gmass of CO2 = 27.73 gmass of H2O = 9.080 g

Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.