We are given a white powder, glycine, with the general formula $C_xH_yO_zN_t$. 1.5g of glycine is reacted with excess sodium hypobromite ($NaBrO$). The escaping gases are passed through two bubblers. The first bubbler, containing sulfuric acid, increases in mass by 0.9g. The second bubbler, containing potash, increases in mass by 1.76g. The nitrogen gas ($N_2$) formed is collected by water displacement, occupying a volume of 225 $cm^3$ at 0.5 $^\circ C$ and a pressure of 75.5 cm of mercury. The molar mass of glycine is given as 75 g/mol. We need to determine the empirical formula of glycine and the mass of sodium bromide ($NaBr$) formed.

1. Problem Description

We are given a white powder, glycine, with the general formula $C_xH_yO_zN_t$. 1.5g of glycine is reacted with excess sodium hypobromite ($NaBrO$). The escaping gases are passed through two bubblers. The first bubbler, containing sulfuric acid, increases in mass by 0.9g. The second bubbler, containing potash, increases in mass by 1.76g. The nitrogen gas ($N_2$) formed is collected by water displacement, occupying a volume of 225 $cm^3$ at 0.5 $^\circ C$ and a pressure of 75.5 cm of mercury.

The molar mass of glycine is given as 75 g/mol. We need to determine the empirical formula of glycine and the mass of sodium bromide ($NaBr$) formed.

2. Solution Steps

First, let's calculate the moles of nitrogen gas ($N_2$) formed using the ideal gas law. We need to convert the given values to standard units.

Temperature: $T = 0.5 ^\circ C = 0.5 + 273.15 = 273.65 K$

Pressure: $P = 75.5 cmHg$. 1 atm is 76 cmHg, so $P = 75.5/76 atm \approx 0.9934 atm$.

Volume: $V = 225 cm^3 = 0.225 L$.

The ideal gas constant $R = 0.0821 L \cdot atm / (mol \cdot K)$.

$PV = nRT$, so $n = \frac{PV}{RT} = \frac{0.9934 \times 0.225}{0.0821 \times 273.65} \approx 0.00994 mol$ of $N_2$.

Since there are $t$ nitrogen atoms in glycine, 1.5g of glycine generates 0.00994 mol of $N_2$. So, $t$ moles of $N$ atoms per mole of glycine are present, leading to $0.00994 \times 2 = 0.01988 \approx 0.02 mol N$ per 1.5 g glycine.

Since the molar mass of glycine is 75 g/mol, the number of moles of glycine in 1.5 g is $1.5/75 = 0.02 mol$.

Therefore, the number of moles of $N$ atoms per mole of glycine is $0.02/0.02 = 1$. Hence, $t=1$.

The first bubbler absorbs water from the reaction. The increase in mass is 0.9g, which is the mass of water ($H_2O$) absorbed. So, moles of $H_2O = 0.9/18 = 0.05 mol$. Since we started with 0.02 mol of glycine, this implies $0.05/0.02 = 2.5$ moles of $H_2O$ per mole of glycine. $y/2 = 2.5$. Thus, $y=5$, so there are 5 hydrogen atoms in glycine.

The second bubbler absorbs carbon dioxide ($CO_2$). The increase in mass is 1.76g, which is the mass of $CO_2$. Moles of $CO_2 = 1.76/44 = 0.04 mol$.

Thus, $0.04/0.02 = 2$ moles of $CO_2$ are produced per mole of glycine. Hence $x=2$, so there are 2 carbon atoms in glycine.

The empirical formula is now $C_2H_5O_zN$. The molar mass is $2(12) + 5(1) + z(16) + 1(14) = 24 + 5 + 16z + 14 = 43 + 16z = 75$.

Then $16z = 75 - 43 = 32$, so $z=2$. Therefore, the empirical formula is $C_2H_5O_2N$.

The reaction of glycine with sodium hypobromite ($NaBrO$) is complex and not fully specified in the problem. We are only given that $NaBr$ is formed. The reaction consumes $NaBrO$ and forms $NaBr$. We can't determine the amount of $NaBr$ produced without knowing the balanced chemical equation.

However, since this is an exercise with only the information provided, we are probably meant to ignore the details of the reaction.

The question is asking for mass of NaBr formed. The atomic mass of Na is 23 and the atomic mass of Br is 80, so the molar mass of NaBr is $23+80 = 103 g/mol$.

Since this question is not well defined given the provided context, I can't estimate the mass of sodium bromide ($NaBr$) formed.

We require more information such as the balanced chemical equation for the reaction.

3. Final Answer

The empirical formula of glycine is $C_2H_5O_2N$.

It is not possible to determine the mass of sodium bromide ($NaBr$) formed with the provided information.