A compound,KBrOx, is analyzed and found to contain 52.92% Br by mass. What is the value of x?
Find the RMM (R) of KBrOx by adding the relative atomic mass (RAM) of the ingredients. Keep the x in the formula.
The RAM of Bromine is 79.9,
so
R=39+79.9+16x (approximate)
79.9/R = 52.92/100
Solve for x and round to the nearest integer.
To determine the value of x in the compound KBrOx, we need to analyze the information provided. It states that the compound contains 52.92% Br by mass.
To calculate the value of x, we can assume 100 grams of the compound. This means that 52.92 grams of the compound are Br, and the remaining mass is due to K (potassium) and O (oxygen).
To find the mass of K, we can subtract the mass of Br from the total mass:
100 g - 52.92 g = 47.08 g
Since the compound only contains K, Br, and O, the remaining mass must be due to oxygen. Therefore, the mass of oxygen is 47.08 g.
Next, we need to determine the molar masses of K, Br, and O:
K: 39.10 g/mol
Br: 79.90 g/mol
O: 16.00 g/mol
To find the value of x, we can calculate the moles of each element present and determine their ratio. The ratio of the moles for each element should be the same as the subscripts in the compound formula.
The moles of K can be calculated by dividing the mass of K by its molar mass:
47.08 g K / 39.10 g/mol = 1.203 mol K
The moles of Br can be calculated in a similar way:
52.92 g Br / 79.90 g/mol = 0.662 mol Br
The moles of O can be calculated as:
47.08 g O / 16.00 g/mol = 2.942 mol O
Based on the molar ratios, we can divide each mole quantity by the smallest number of moles (0.662 mol Br) to obtain whole number ratios:
K: 1.203 mol / 0.662 mol ≈ 1.815
Br: 0.662 mol / 0.662 mol = 1
O: 2.942 mol / 0.662 mol ≈ 4.446
Rounding these ratios to the nearest whole number gives us the simplest ratio of KBrOx, which is KBrO4.
Thus, the value of x in the compound KBrOx is 4.