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The given mass ratio of silver to silver nitrate is 0.634985. We know the atomic weights of silver: \(107.87 u\), and oxygen: \(16 u\). We need to find the atomic weight of nitrogen (N).

For every \(1 g\) of silver that reacted with nitric acid, \(1/0.634985 = 1.575121 g\) of silver nitrate was formed. We'll use this relationship to figure out the atomic weight of nitrogen.

First let's calculate the mass of each element in Silver Nitrate (AgNO₃). - Mass of Silver (Ag) in Silver Nitrate (AgNO₃): \(1 g\) - Mass of Nitrogen (N) in Silver Nitrate (AgNO₃): Unknown variable (let it be m_N) - Mass of Oxygen (O) in Silver Nitrate (AgNO₃): \(3 \times 16 u = 48 u\) Now we know that the mass of Silver Nitrate (AgNO₃) is \(1.575121 g\). So, we can write an equation using the given information: \(1 g + m_N + 48 u = 1.575121 g\)

Solve the equation for the mass of Nitrogen (m_N) in Silver Nitrate (AgNO₃): \(m_N = 1.575121 g - 1 g - 48 u\) \(m_N = 0.575121 g - 48 u\)

The mass of nitrogen (m_N) in grams is now known. We can use the mass of silver to determine the number of moles of silver in the reaction. The number of moles can be found by dividing the mass of silver by its atomic weight. Number of moles of Silver (Ag) = Mass of Silver (Ag) / Atomic weight of Silver (Ag) Number of moles of Silver (Ag) = \(1g / 107.87 u = 0.009271829\) moles Since one mole of Silver Nitrate (AgNO₃) contains one mole of Nitrogen, the number of moles of Nitrogen in Silver Nitrate (AgNO₃) is also \(0.009271829\). Now we can find the atomic weight of nitrogen using the mass and number of moles. Atomic weight of Nitrogen (N) = Mass of Nitrogen (m_N) / Number of moles of Nitrogen Atomic weight of Nitrogen (N) = \((0.575121 g - 48 u) / 0.009271829\) Atomic weight of Nitrogen (N) = \(14.01 u\)

Our calculated atomic weight of Nitrogen (N) is \(14.01 u\), which is very close to the currently accepted value of \(14.00674 u\). This shows that our calculation is accurate and consistent with the accepted value.