A compound is analyzed and determined to be 30.4% nitrogen and 69.6% oxygen. What is the empirical formula of the compound?

NO2

Explanation:

The empirical formula represents the simplest whole number ratio of the elements in a compound. According to the question;

30.4% represents 30.4g of nitrogen

69.6% represents 69.6g of oxygen

First, we convert each mass value to mole by dividing by their respective molar masses as follows:

N = 30.4/14 = 2.17mol

O = 69.6/16 = 4.35mol

Next, we divide each mole value by the smallest (2.17)

N = 2.17/2.17 = 1

O = 4.35/2.17 = 2.0046

Approximately, the ratio of N to O is 1:2, hence, the empirical formula is NO2

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.