Compound x has a molar mass of 230.29 g-mol element mass % carbon 62.58% hydrogen 9.63% oxygen 27.79% Write the molecular formula of x. and the following composition:

Q: Calculate the molar mass for H2SO4

A: Given, Compound - H2SO4 (sulfuric acid) Molar mass of H2SO4 (sulfuric acid) =? Molar mass: Molar…

Q: Compound X has a molar mass of 176.12 g mol and the following composition: element mass % carbon…

A: Answer:Molecular formula of the compound expresses the number of moles of atoms of each type that…

Q: A compound consisting of B, N, and H undergoes elemental analysis. The % composition by mass is…

A: Given,Percent of B in the compound = 40.28 %Percent of N in the compound = 52.20 %Percent of H in…

Q: Which of the following could only be an empirical formula? N2H4 N2O4 .N2H6. N2O5

A: Empirical formula of a compound is defined as the formula of a chemical compound in which atoms are…

Q: The number of moles in 37.80 g of C2H6O (molar mass = 46.1 g/mol) is?

A: Given information,       Mass of C2H6O is 37.80 g.     Molar mass of C2H6O is 46.1 g/mol.

Q: Compound X has a molar mass of 134.12 g-mol and the following composition: element mass % carbon…

A: Answer: Molecular formula is the way of representing the molecule in which chemical formula tells us…

Q: Which of the following could be an empirical formula?

A: Empirical formula of a compound written with the smallest possible integers.

Q: ruthenium chloride contains 48.72% ruthenium and 51.27% chlorine by mass. what is the empirical…

A: mass % of ruthenium (Ru) = 48.72 % = 48.72 gmass % of chlorine (Cl) = 51.27 % = 51.27 gEmpirical…

Q: Determine the molecular formula for the compound with a molar mass of 60.10g/mol and the following…

A:

Q: Molar mas of

A:

Q: An unknown substance has a mass percent composition of C :49.48% H:5.19% , N:28.85%, O: 16.48% with…

A: To calculate molecular formula, empirical formula needs to be calculated. Percent composition of the…

Q: 4. Determine the molecular formula of the compound with an empirical formula of CH and a formula…

A:

Q: Calculate the mass percent composition of carbon in carbon containing compound. CH4

A: The mass percent composition of carbon in carbon containing compound, CH4 is to be calculated.

Q: As a result of burning analysis of a caffeine sample 0,624 g C, 0,065 g H, 0,364 g N ve 0,208 g O…

A: Number of moles = mass / Molar mass

Q: What is the empirical formula of a compound of carbon, hydrogen, and Oxygen that contains 51.56% and…

A: Given:: %C = 51.56%  %H = 14.09 % then, empirical formula = ?

Q: What is the empirical formula for a sample containing 81.70% carbon and 18.29% hydrogen?

A: Sample containing 81.70% carbon and 18.29% hydrogen .

Q: Calculate the empirical formula for a compound using the element mass percent composition of 75.69%…

A: Let the mass of the unknown compound be 100 g.So,mass of carbon in the compound = 75.69 gmass of…

Q: What is the percent composition of silver nitrate, AgNO3?

A: % composition = n× atomic mass×100/molar mass n= number of atoms

Q: Compound X has a molar mass of 134.12 g-mol and the following composition: element 0 carbon mass %…

A:

Q: 37. Medical implants and high-quality jewelry items for body piercings are frequently made of a…

A: 37. G23Ti or surgical-grade titanium has 64.39 % titanium, 24.19 % aluminum and 11.42 % vanadium…

Q: -1 Compound X has a molar mass of 135.12 g mol element carbon mass % 44.44% hydrogen 3.73% nitrogen…

A: Molecular formula: In a molecular formula we describe the number of atoms of all the elements…

Q: Determine the molecular formula of a compound that has a molar mass of 183.2 g/mol and an empirical…

A: Molecular formula: In a molecular formula we describe the number of atoms of all the elements…

Q: Determine the mass of each element present in a 125.0-gram sample of this chemical.

A:

Q: ste and smell) natural flavor based on its elementa tion. C 58.80%, H 9.87%, O 31.33% 0₁ H9.791.958…

A: If mass of compound is 100 g. Mass of C = 58.80 g Moles of C = 58.80/12 =4.90 mol Mass of H = 9.87 g…

Q: How many moles of (NH4)3PO4 are contained in 0.917 g of this compound?

A: Given that: Mass of (NH4)3PO4 = 0.917g Moles of (NH4)3PO4 = ?

Q: 89% copper oxide What is the

A: Given percentage of Copper Cu=89% Given percentage of oxygen=(100-89)=11%

Q: Q.26. The Elemental analysis of a compound sample informed that it contains 7.83 g of oxygen, 4.35 g…

A: Given  Mass of oxygen = 7.83 gram Molar mass of oxygen = 16 gram /mole  Therefore number of mole of…

Q: Compound X contains 82.4 % nitrogen and 17.6 % hydrogen. What is the empirical formula?

A: Empirical formula is the simplest positive integer ratio of atoms present in a compound.

Q: Calculate the empirical formula for the compound found to have the following elemental composition…

A: Empirical formula is the simplest ratio of the number of elements present in the given molecular…

Q: lar mass of 316.29 g-mol and the following compos mass % 39.17% 60.83% rmula of X.

A: %mass of P = [ ( x • mass of P ) / ( molar mass of Compound X) ] • 100  39.17 / 100 = ( x • 31gmol-1…

Q: A compound has the formula ZnGa2O4. What is the percent composition?

A: In this question we want to determine the value of percent composition of Zn, Ga and O for these…

Q: What is the percent composition by mass (to one decimal place) of Oxygen and hydrogen in 1 mol of…

A: The mass of 1 mol of atoms is called atomic mass For example :-  Atomic mass of H = 1.0 g/mol Atomic…

Q: Compound X has a molar mass of 230.29 g-mol element 1 carbon mass % 62.58% hydrogen 9.63% 27.79%…

A: Molecular formula describes the number of atoms of all the elements present in the molecule by using…

Q: A compound having an approximate molar mass of 165-170g has the following percent composition by…

A: From percentage composition we can determine empirical and molecular formula. Empirical Formula =…

Q: Determine the empirical formula for a compound that is 35.00% nitrogen, 5.05% hydrogen and 59.95%…

A:

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.